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LUCKY_DIMON [66]
3 years ago
8

A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

gnitude.) _________ N
Physics
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter a=1.6m/sec^2

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So F=70\times 1.6=112 N

So external force will be 112 N

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HELP PLEEAAAASSSEEEEEEE What is the definition of net force?
aleksley [76]

Answer:

the sum of all force being applied to an object.

Explanation:

8 0
3 years ago
Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7
inysia [295]

Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

2.063d = 10.63  

d = 5.15m  

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

4 0
3 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ ௠at a d
Angelina_Jolie [31]

A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ ௠at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?

If the distance from the source is doubled. The amplitude of the magnetic field is smaller 4 times.

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3 years ago
What would happen if you place 2 Plane Mirrors parallel to each other?
e-lub [12.9K]

Answer:

No image will be observed.

Explanation:

Images that are created by mirrors are virtual images. This virtual image can only be seen by an observer. In this case, an infinite number of images or no image will be created here as both will be reflecting their own images. Light will continuously bounce back and forth reflecting the same image.

6 0
3 years ago
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