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3 years ago

An object with a mass of 2000 kg accelerates 8.3 m/s2 when an unknown force is applied to it. What is the amount of the force?

Physics

1 answer:

exis [7]3 years ago

6 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 2000 × 8.3

We have the final answer as

Hope this helps you

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An object that is completely or partially submerged in fluid experience an upward force called?

romanna [79]

The buoyant force or upward buoyancy force

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2 years ago

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

<u>Given:</u>

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

3 0

2 years ago

A light ray is incident from air into glass (ng = 1.52) then onto water (nw= 1.33). The wavelength of light in air (na= 1) is ai

gregori [183]

The answer is is b hope this helped

3 0

3 years ago

11. Velocity-time graph for the motion of an object in a straight path is a straight line parallel to the time axis

Vilka [71]

a) uniform velocity

b) zero or no acceleration

c) (see picture)

EXPLANATION:

(see picture)

8 0

2 years ago

31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o

coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

$y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ is the final height= 0

$y_{0}$ is the initial height= 1.5 m

$v_{0y}$ is the component of the initial speed in the vertical direction = 0 m/s

t: is the time =?

g: is the gravity = 9.81 m/s²

$0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}$

By solving the above equation for t we have:

$t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s$

Hence, the ball will stay 0.55 seconds in the air.

B) We can find the distance traveled by the ball as follows:

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

Where:

a: is the acceleration in the horizontal direction = 0

$x_{f}$ is the final position =?

$x_{0}$ is the initial position = 0

$v_{0x}$ is the component of the initial speed in the horizontal direction = 45 m/s

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

$x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m$

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

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2 years ago