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vagabundo [1.1K]
3 years ago
8

The motion of a car on a position time graph is represented with a horizontal tine What does this indicate about the car's motio

n?
Physics
1 answer:
GalinKa [24]3 years ago
8 0

Answer:

Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.

The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.

A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.

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tiny-mole [99]

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B

Explanation:

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3 years ago
If you drive at 100 km/hr for six hours how far will you go
zhuklara [117]

600km/h


100 times 6 = 600

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Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball
denpristay [2]

Answer:

1:4

Explanation:

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KE=\dfrac{1}{2}mv^2

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

3 0
3 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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3 years ago
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Answer:

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Explanation

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