A metal ball sits motionless on a flat surface. Which of these would make the ball move?
A. The force of gravity becomes less.
B. The force of gravity becomes greater.
C. Two equal horizontal opposing forces act upon the ball.
D. Two unequal horizontal opposing forces act upon the ball.
Answer:
D
Explanation:
Unbalanced forces move stuff. Gravity would only increase/decrease movement if the object was already in motion.
Answer:
b
Explanation:
1. What is the chemical name of the following chemical symbols?
Be
Nel
Mg
Na
Answer:
Na sodium
Mg magnesium
Be beryllium
Explanation:
Nel is not any element it is wrong
anyone to assist me on it ...especial page7 and 8
Answer:
i needed points it was an emergency sorry
Explanation:
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
In which medium does the light move faster, water or diamond?
The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be
This question is incomplete
Complete Question
m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.
a) if the system is released from rest what will be its acceleration
Answer:
0.7 m/s²
Explanation:
The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.
(a) if the system is released from rest what will be its acceleration
g = acceleration due to gravity = 9.81 m/s²
Coefficient of Kinetic Friction = μk = 0.30
m1 = 10kg
m2 = 4.0kg
The formula to solve question a is given as:
a = acceleration at rest
m2g- μk m1g = (m1+ m2) a
Making a the subject of the formula:
a = (m2g- μk×m1g )/(m1+ m2)
a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)
a = 0.7 m/s²
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!
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
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
The volume of water in a water bottle, is about 398
g
cm
km/hr
Kg
g/mL
ml
km
m/s
Answer:
milliliters (ml)
Explanation:
millileters is the correct measurement for liquids
3 For this force system the equivalent system at P is ___________ A FRP 40 lb along x dir and MRP 60 ft lbB FRP 0 lb and MRP 30 ft lbC FRP 30 lb along y dir and MRP 30 ft lbS FRP 40 lb along x dir and MRP 30 ft lb
This question is incomplete, the complete question is;
For this force system the equivalent system at P is ___________
A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft.lb
B) FRP = 0 lb and MRP = +30 ft.lb
C) FRP 30 lb (along +y-dir.) and MRP = -30 ft.lb
D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb
Answer:
D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb
Explanation:
From the figure in the image i uploaded along this answer;
FRP = ( 40 lb i + 30 lb j ) + [30 lb (-j)]
Where i and j are the unit vectors along X & Y axis respectively.
So, FRP = 40 lb i
that is, FRP = 40 lb along +X direction
MRP = [ 30 lb x ( 1 ' + 1' ) ] +( -30 lb x 1 ' )
= (30 lb x 2 ' )- 30 lb ft
= 60 lb ft - 30 lb ft
= 30 lb ft
Therefore option(D) is correct
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 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
gold has a density of 19.32g/cm3. if you have a 25 cm3 sample of gold what is the mass of the sample
Answer:
ggggggggggggggggggggggggggggg
Explanation:
Answer:
The volume of the sample of gold is
16.51 [tex]cm^{3}[/tex]
Explanation:
The formula for density is:
D= [tex]\frac{M}{V}[/tex].
where:
D is density,
M is mass, and
V is volume.
Rearrange the density formula to isolate volume.
V= [tex]\frac{M}{D}[/tex]
V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]
V= 318.97∅ × [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.
V= 16.51 cm³ Au.
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 7.63 m/s in 3.94 s. What is the magnitude of the linear impulse experienced by a 73.7 kg passenger in the car during this time? Submit Answer Tries 0/20 What is the average force experienced by the passenger?
Answer:
1. p = 562.3 kg*m/s
2. F = 142.7 N
Explanation:
1. The linear impulse (p) is given by:
[tex] p = mv [/tex]
Where:
m: is the passenger's mass = 73.7 kg
v: is the speed = 7.63 m/s
[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]
Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.
2. The average force can be calculated using the following equation:
[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]
Therefore, the average force experienced by the passenger is 142.7 N.
I hope it helps you!
calculating light in physics
Find the angle between the two unitless vectors: F1 = 8.92 i + 17.37 j F2 = 12.44 i + 7.11 j Answer in degrees, and to the fourth decimal place.
Answer:
θ = 33.0705°
Explanation:
The angle between the two vectors is given by the formula;
Cos θ = (F1 • F2)/(|F1| × |F2|)
We are given;
F1 = 8.92i + 17.37j
F2 = 12.44i + 7.11j
Thus;
Cos θ = [(8.92i + 17.37j) • (12.44i + 7.11j)]/[√(8.92² + 17.37²) × √(12.44² + 7.11²)]
Cos θ = (110.9648 + 123.5007)/(19.5265 × 14.3285)
Cos θ = 0.8380
θ = cos^(-1) 0.8380
θ = 33.0705°
which equation should be used to find speed
Answer:
The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).
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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
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.
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
An object is accelerating if it is moving?
9514 1404 393
Answer:
Not Necessarily
Explanation:
If the object is changing speed or direction, then it is accelerating. If it is maintaining the same speed and direction, it is not accelerating.
what happens to the temperature of water as time elapses? IF YOU ANSWER IT I WILL MARK YOU A BRAINLEST ANSWER
Answer:
I think it will get colder
Explanation:
Answer:
The water molecules go faster as it gets colder they go slower
Explanation:
trust me thats the answer
During which process of the water cycle does water change from a gas to liquad
At time t = 0 the point at x = 0 has velocity v0 and displacement y0. The phase constant φ is given by tanφ =:
This question is incomplete, the complete question is;
The displacement of a string carrying a traveling sinusoidal wave is given by y(x,t)=ymsin(kx - ωt -φ) .
At time t = 0, the point at x = 0 has velocity v₀ and displacement y₀.
The phase constant φ is given by tanφ =:
A) ωv₀ /y₀
B) ωv₀ y₀
C) v₀ /ωy₀
D) y₀ /ωv₀
E) ωy₀ /v₀
Answer:
E) ωy₀ /v₀
Explanation:
Given that;
displacement of a wave is; y(x,t) = ym sin (kx - ωt - φ)
we differentiate the given equation with respect to time
d/dt (y(x,t)) = d/dt(ym sin(kx - ωt - φ) )
v(0,0)) = -ym ωcos (k(0) - ω(0) - φ) )
v₀ = -ym ωcos (-φ) ......... lets leave thisas equ 1
At t = 0, x = 0
the displacement of the wave is
y(0,0) = ym sin (k(0) - ω(0) - φ)
y₀ = ym sin(-φ) ..............let this be equ 2
y₀/v₀ = (ym sin(-φ)) / (-ym ωcos (-φ)) = ( -ym sin(φ)) / (-ym ωcos (φ))
(tanφ)/ω = y₀/v₀
tanφ = y₀ω/v₀
therefore the required value is y₀ω/v₀
option (E).
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!
A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.
Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.
Answer:
The value is [tex]H = 18*10^{2} \ Atom / sec [/tex]
Explanation:
From the question we are told that
The atom fraction of metal A at point G is [tex] A = 0.30 \ m[/tex]
The atom fraction of metal A at a distance 5000nm from G is [tex]A_2 = 0.35[/tex]
The number of atoms per [tex]m^3[/tex] is [tex]N_h = 9 * 10^{28}[/tex]
The diffusion coefficient is [tex]D = 2* 10^{-14 } m^2/s[/tex]
Generally of the concentration of atoms of metal A at G is
[tex] N_A = A * N_h [/tex]
=> [tex] N_A = 0.3 * 9 * 10^{28}[/tex]
=> [tex] N_A = 2.7 * 10^{28} 2.7 atoms/m^3[/tex]
Generally of the concentration of atoms of metal A at a distance 5000nm from G is
[tex]D = 0.35 *9 * 10^{28}[/tex]
=> [tex]D = 3.15 * 10^{28} \ atoms / m^3[/tex]
The concentration gradient is mathematically represented as
[tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = 9 * 10^{20} / m^4[/tex]
Generally the flux of the atoms per unit area according to Fick's Law is mathematically represented as
[tex]J = -D* \frac{d N_A}{dx}[/tex]
=> [tex]J = -2* 10^{-14 * 9 * 10^{20} [/tex]
=> [tex] J = 18*10^{6}\ atoms\ crossing\ /m^2 s [/tex]
Generally if the cross-section area is [tex] a = 1 cm^2 = 10^{-4} \ m^2[/tex]
Generally the number of atom crossing the above area per second is mathematically is
[tex]H = 18*10^{6} * 10^{-4} [/tex]
=> [tex]H = 18*10^{2} \ Atom / sec [/tex]
A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what
is the value of the current?
A. 3.0 A
B 5.0 A
C. 11 A
D. 15 A
E. 20A
Answer:
B 5.0 A .
Explanation:
Hello.
In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:
[tex]I=\frac{Q}{t}[/tex]
Then, for the given data, we obtain:
[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]
Therefore, answer is B 5.0 A .
Best regards!
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
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