A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person

Answers

Answer 1

Answer:

20000

Explanation:

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000


Related Questions

calculating light in physics

Answers

Formula: c = f where:
c = the speed of light = 300,000 km/s or 3.0 x 108 m/s.
= the wavelength of light, usually measured in meters or Ångströms (1 Å = 10-10 m)
f = the frequency at which light waves pass by, measured in units of per seconds (1/s).

If the body with a mass of 4kg is moved by a force of 20 N, what is the rate of its acceleration?

Answers

Answer:

The answer is 5 m/s²

Explanation:

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/s²

Hope this helps you

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

Answers

The answer is c i think

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.

Answers

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°

What energy transformation takes place when you stretch a bungee cord?

Answers

Answer:

potential energy

Explanation:

Please provide an explanation.

Thank you!!

Answers

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 kN

Explanation:

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

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.

Answers

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 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 !

Answers

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

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?

Answers

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.

which equation should be used to find speed

Answers

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).

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,

Answers

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

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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.

Answers

Answer:

the answer is C

Explanation:

you said its c

anyone to assist me on it ...especial page7 and 8

Answers

Answer:

i needed points it was an emergency sorry

Explanation:

During which process of the water cycle does water change from a gas to liquad

Answers

Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.

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.

Answers

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.

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

Answers

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    

1. What is the chemical name of the following chemical symbols?
Be
Nel
Mg
Na

Answers

Answer:

Na sodium

Mg magnesium

Be beryllium

Explanation:

Nel is not any element it is wrong

gold has a density of 19.32g/cm3. if you have a 25 cm3 sample of gold what is the mass of the sample​

Answers

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.

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

Answers

Answer:

A

Explanation:

hope this helps

The answer is A thank me later

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

Answers

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. ​

Answers

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 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

Answers

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

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

Answers

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!

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?

Answers

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!

An object is accelerating if it is moving?


Answers

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.

Please provide explanation!!!
Thank you.

Answers

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

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.

Answers

Answer:

D

Explanation:

Unbalanced forces move stuff. Gravity would only increase/decrease movement if the object was already in motion.

Answer:

b

Explanation:

What is the correct answer?

Answers

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

The volume of water in a water bottle, is about 398
g
cm
km/hr
Kg
g/mL
ml
km
m/s

Answers

Answer:

milliliters (ml)

Explanation:

millileters is the correct measurement for liquids

what happens to the temperature of water as time elapses? IF YOU ANSWER IT I WILL MARK YOU A BRAINLEST ANSWER​

Answers

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

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