Answer:
A force can cause an object to accelerate, slow down, remain in place, or change shape. The unit of measure for force is the newton which is abbreviated as "N". One newton is the force needed to accelerate one gram of mass by one centimeter per second squared. Other units of force include the dyne and the pound-force.
What is the correct answer?
Answer:
2156 N
Explanation:
Data obtained from the question include:
Mass of satellite (m) = 220 Kg
Force (F) of gravity =?
The force of gravity exerted on the satellite on the surface of the earth can be obtained by using the following formula:
Force (F) of gravity = mass (m) × acceleration due to gravity (g)
F = mg
Mass of satellite (m) = 220 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) of gravity =?
F = mg
F = 220 × 9.8
F = 2156 N
Thus, the force of gravity exerted on the satellite on the surface of the earth is 2156 N
A kangaroo is traveling at a velocity of 14.5 m/s for 7.5 seconds. The kangaroo’s force generated by its legs is 100 N. What is the mass of the kangaroo
A. 59kg
B. 63.28 N
C. 51.81 N
D. 51.81 kg
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3
4
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Dressing appropriately for exercise includes
A. wearing the same clothes for all exercises
B. choosing dark colored clothing when exercising at night
C. wearing sunscreen when exercising outside
D. making sure you wear the best brand-name clothes
Please select the best answer from the choices provided.
A
B.
C.
D.
The answer is C.
Answer:
the answer is C
Explanation:
you said its c
100 POINTS.
Please provide explanation.
Thank you
Answer:
(a) 0.829 m/s
(b) 3.27 m/s
(c) 0.000153 m²
55.8%
Explanation:
(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)
Q = vA
(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)
v = 0.829 m/s
(b) Use Bernoulli equation. Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head. Choose 0 elevation to be at point 1.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)
1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa
v = 3.27 m/s
(c) Flow rate is constant.
Q = vA
(0.001 m³ / 2.00 s) = (3.27 m/s) A
A = 0.000153 m²
Flow rate is proportional to the pressure difference and the radius raised to the fourth power.
Q ∝ ΔP r⁴
Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴
Q₂/Q₁ = (1.120) (0.840)⁴
Q₂/Q₁ = 0.558
The flow decreases to 55.8% of the original value.
Answer:
Explanation:
Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed. As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.
In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.
The new flow rate = (1.12)*(0.84)^4
=0.5576 or 55.76% of the original flow rate
The bandgap of InP semiconductor laser is 1.0 eV. The effective mass of the valence band is ½ of the effective mass of the conduction band. Assuming that electron hole recombination transition occurs at 0.03 eV above the bandgap, calculate the wavelength of this transition.
Answer: the wavelength of this transition is 1.2039 um
Explanation:
Given that;
the energy level between the transitioning energy gap Eg = 1.0 + 0.03 = 1.03 eV
we know that λ = 1.24 / Eg
so we substitute our Eg into the above equation
λ = 1.24 / 1.03
λ = 1.2039 um
therefore the wavelength of this transition is 1.2039 um
Please provide an explanation.
Thank you!!
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kNb. Left support = 36 kN Right support = 29 kNc. Left support force will decrease Right support force will increase.d. Right support = 43 kNExplanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
What type of observation is made through interviewing people’s
Answer:
Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.
A "lovesick" individual wants to throw a bag of candy and love notes into the open window of their significant other’s bedroom 10.0 m above. Assuming it just reaches the window, they throw the gift at 60.0o to the ground: At what velocity should they throw the bag? How far from the house are they standing when they throw the bag? (Answer: A. 16.2m/s B. 11.5m)
Answer:
Explanation:
Let the velocity be v .
vertical component of the velocity = v sin 60 = √3 v /2
it reaches maximum height of 10 m .
v² = 2 gh
( √3 v/2 )² = 2 x 9.8 x 10
3 v² = 196 x 4
v² = 65.33 x 4
v = 16.2 m /s
Let time taken to reach height of 10 m
v = u - gt
v sin 60 = 9.8 t
16.2 x √3 /2 = 9.8 t
t = 1.43 s
horizontal distance covered = v cos 60 x t
16.2 x .5 x 1.43 = 11 .5 m
If the body with a mass of 4kg is moved by a force of 20 N, what is the rate of its acceleration?
Answer:
The answer is 5 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]acceleration = \frac{force}{mass} \\[/tex]
From the question
force = 20 N
mass = 4 kg
We have
[tex]a = \frac{20}{4} \\ [/tex]
We have the final answer as
5 m/s²Hope this helps you
PLEASE PROVIDE AN EXPLANATION!!
THANK YOU.
Answer:
(a) 3.43 m/s
(b) 3.43 m/s
(c) 95.8 kPa
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
where P is pressure (either absolute or gauge), ρ is density, v is velocity, g is acceleration due to gravity, and h is elevation.
(a) Let's choose point 1 at the surface of the fluid in the container, and point 2 at point Z at the exit of the tube. I'll say 0 elevation is at point Z, and I'll use gauge pressure.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 Pa + ½ ρ (0 m/s)² + ρ (9.8 m/s²) (0.60 m) = 0 Pa + ½ ρ v² + 0
ρ (9.8 m/s²) (0.60 m) = ½ ρ v²
5.88 m²/s² = ½ v²
v = 3.43 m/s
(b) The tube's cross section is constant, so the fluid's speed is the same at all points in the tube. v = 3.43 m/s.
(c) Use Bernoulli equation again, choosing point 2 to be at Y. I'll say 0 elevation is at the surface of the fluid, and again use gauge pressure.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + 0 + 0 = P + ½ (700 kg/m³) (3.43 m/s)² + (700 kg/m³) (9.8 m/s²) (0.20 m)
0 = P + 4116 Pa + 1372 Pa
P = -5488 Pa
The gauge pressure is -5488 Pa, so the absolute pressure is 101,300 Pa + -5488 Pa = 95812 Pa, or approximately 95.8 kPa.
how are s waves and p waves simuliar?
A.they shake the ground
B.they travel through liquids
C. they arrive at the same time
D.they shake the ground from side to side
Answer:
A
Explanation:
hope this helps
Hi, Please help.. I have assignments due tonight and I need someone to help me when a question I have generally..
Okay so if Density = Mass divided by Volume and I put that information into a calculator it comes out as
ex. 1.938773646 how do I make it look like something like this 1.4?
Answer:
did you tried putting it in standard form
Answer:
It may help to round all of the given numbers up to at least 1 or 2 decimal points or you could round up the number you get to 1 or 2 decimal places. For example, for this question round up your answer to 1.9 or 1.94
Explanation:
hope this helps!!
During which process of the water cycle does water change from a gas to liquad
A car is traveling south at 8.77 m/s. It then begins a uniform acceleration until it reaches a velocity of 47.8 m/s over a period of 3.84s. What is the car's acceleration?
Please help !
Answer:
The acceleration of the car is 10.16m/s²
Explanation:
Given parameters:
Initial velocity = 8.77m/s
Final velocity = 47.8m/s
Time duration = 3.84s
Unknown:
Acceleration of the car = ?
Solution:
To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;
Acceleration = [tex]\frac{V - U}{T}[/tex]
V is the final velocity
U is the initial velocity
T is the time taken
Input the parameters and solve for acceleration;
Acceleration = [tex]\frac{47.8 - 8.77}{3.84}[/tex] = 10.16m/s²
The acceleration of the car is 10.16m/s²
The diagram shows two forces acting on the dog. What are these two forces
Answer:
kenietic and potential i guess
Explanation:
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement
Explanation:
Let east = E, and, west = opposite to east = - E.
Here, displacement:
=> 2m east + 4m west + 1m east
=> 2E + 4(-E) + 1E
=> 2E - 4E + 1E
=> - 1E
=> 1(-E)
=> 1m west
And, distance,
=> 2m + 4m + 1m = 7m
The distance of a person is 7 m and the displacement of the person is 1m west.
To find the distance and displacement, the given values are,
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.
What is the distance and the displacement?Displacement:
The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.Distance:
The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.Let us consider East = E and west = opposite to east = - E.
Calculating the displacement:
= 2m east + 4m west + 1m east
= 2E + 4(-E) + 1E
= 2E - 4E + 1E
= - 1E
= 1(-E)
= 1m west.
The displacement is 1m west.
Now calculating the distance,
= 2m + 4m + 1m
= 7m
The distance is 7m.
Thus, the displacement and the distance is found as 1 m west and 7m.
Learn more about distance and displacement,
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An object is dropped from rest and falls freely 20 m to Earth. When is the speed of the object 9.8 m/s?
At the end of the first second of its fall
At the end of the first second of its fall
During the entire time of its fall
During the entire time of its fall
At no time is the speed 9.8 m/s
At no time is the speed 9.8 m/s
During the entire first second of its fall
During the entire first second of its fall
After it has fallen 9.8 meters
Logan is a runner he in 60 seconds he can run 360 m what speed did he travel at
Answer:
hhhhhhhh
Explanation:
A man of weight 60N climbed stairs of height 15m high in 15s. Find the power
of the man,
Two identical items, object 1 and object 2, are dropped from the top of a 50.0m50.0m building. Object 1 is dropped with an initial velocity of 0m/s0m/s, while object 2 is thrown straight downward with an initial velocity of 13.0m/s13.0m/s. What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground
Answer:
Δt = 1.1 s
Explanation:
Given information:
H= 50.0 m
g= 9.8 m/s²
Object 1v₀ = 0[tex]H = \frac{1}{2} * g* t^{2}[/tex]
Solving for t, we get:
t₁= 3.2 s
Object 2v₀ = 13 m/sWe can find the final velocity for the object when it hits the ground, using the following expression:
[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]
Solving for vf, we get:
vf = 33.9 m/s
Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:
[tex]v_{f} = v_{o} + g*t[/tex]
(Assuming the downward direction to be positive).
Solving for t, we get:
t₂ = 2.1 s
So the difference in time when both objects hit the ground, it's simply
Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
Who was the first who traveled to the moon?
NEIL ARMSTRONG WAS THE FIRST MAN WHO TRAVELLED TO THE MOON.
Answer:
On July 20, 1969, Neil Armstrong became the first human to step on the moon.
A single-turn circular loop of radius 9.4 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70degrees below the horizontal north direction. Find the current in the loop.
Answer:
The current in the loop is 10.5 A.
Explanation:
Given that,
Radius = 9.4 cm
Magnetic field = 0.7 G
Angle = 70°
We know that,
The magnetic field due to the current in a loop is
[tex]B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
The magnetic field due to the current is equal to the magnetic field of earth.
[tex]B_{E}=B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
We need to calculate the current in the loop
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}NI}{2r}[/tex]
[tex]I=\dfrac{2rB}{\mu_{0}N}[/tex]
Put the value into the formula
[tex]I=\dfrac{2\times9.4\times10^{-2}\times0.7\times10^{-4}}{4\pi\times10^{-7}\times1}[/tex]
[tex]I=10.5\ A[/tex]
Hence, The current in the loop is 10.5 A.
A child and sled with a combined mass of 53.9 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 5.71 m/s at the bottom, what is the height of the hill
Answer:
1.66m
Explanation:
Using the conservation law
PE = KE
mgh = 1/2mv²
gh = V²/2
g is the acceleration due to gravity = 9.81m/s²
h is the height of the hill
V is the velocity = 5.71m/s
Substitute
9.81h = 5.71²/2
Cross multiply
2×9.81h = 5.71²
19.62h = 32.6041
h = 32.6041/19.62
h = 1.66m
Hence the height of the hill is 1.66m
What energy transformation takes place when you stretch a bungee cord?
Answer:
potential energy
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis. What happens as you move away from the center axis toward the coil? What happens above the coil? Outside the coil? Below the coil?
Answer:
Please find the answer in the explanation
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.
What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.
What happens above the coil?
the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines
Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.
Below the coil?
The needle will move in an opposite direction.
A cup of hot coffee initially at 95 degrees C cools to 80 degrees C in 5 min while sitting in a room of temperature 21 degrees C. Using Newton's law of cooling, determine when the temperature of the coffee will be a nice 50 degrees C.
Answer:
When the temperature of the coffee is 50 °C, the time will be 20.68 mins
Explanation:
Given;
The initial temperature of the coffee T₀ = 95 °C
The temperature of the room = 21°C
Let T be the temperature at time of cooling t in mins
According to Newton's law of cooling;
[tex]\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}} = \int\limits kdt\\\\Log(T-21) =kt + Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453[/tex]
When the temperature is 50 °C, the time t in min is calculated as;
[tex]T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins[/tex]
Therefore, when the temperature of the coffee is 50 °C, the time will be 20.68 mins
In 10 minutes the hot coffee will attain the temperature of 50 degrees Celsius.
Initially the hot cup of coffee at the temperature of 95 degrees Celsius but after 5 minutes its temperature decreases from 95 to 85 degrees Celsius which is 15 degrees Celsius decrease so in other 5 minutes, the temperature decreases to 65 degrees Celsius.
Again after 5 minutes the temperature will further decrease finally the cup of coffee attain the temperature of 50 degrees Celsius so we can conclude that in 10 minutes the hot coffee will gain the 50 degrees Celsius temperature.
Learn more: https://brainly.com/question/18402982
Please provide explanation!!!
Thank you.
Answer:
(a) 102 cm/s
(b) 0.490 cm²
Explanation:
(a) Use Bernoulli equation.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0
½ ρ v₁² + ρgh₁ = ½ ρ v₂²
½ v₁² + gh₁ = ½ v₂²
½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²
v = 102 cm/s
(b) The flow rate is constant.
v₁ A₁ = v₂ A₂
(25.0 cm/s) (2.00 cm²) = (102 cm/s) A
A = 0.490 cm²
The smokestack of a stationary toy tra in consists
of a vertical spring gun chat shoots a steel ball a meter
or so straight into the air-----so straight that the ball
always falls back into the smokescack. Suppose the
train moves at constant speed along the straight track. Do you think the ball will still return to the smoke-
stack if shot from the moving train? What if the train
gains speed along the straight track? What if it moves at a constant speed on a circular track? Discuss why your answers differ,
Answer:
i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
Explanation:
The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
if you are driving 110 km/h along a straight road and you look to the side for 2.0 s , how far do you travel during this inattentive period ? explain.
Explanation:
hope this helps, have a good one :D
Answer:
60.12m
Explanation:
Distance = Velocity x Time
To use this formula we must first convert 110km/h to m/s, which we can do by dividing the value by 3.6:
110/3.6 = 30.56m/s (2dp)
Velocity = 30.56m/s
Time = 2s
Distance = 30.56x2
Distance = 61.12m
You travel 60.12m during this inattentive period.
Hope this helped!