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2 years ago

What is the mass of an object that requires 100N of force in order to accelerate 10m/s2

Physics

1 answer:

iragen [17]2 years ago

3 0

Explanation:

Hey there!

Given;

Force (f) = 100 N

Acceleration (a) = 10m/s^2.

Mass (m)=?

We have;

Force = mass * acceleration.

So,

100 = m * 10

Or, 100 = 10m

Or, m= 100/10

Therefore, mass is 10kg.

Hope it helps....

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padilas [110]

Answer:

D. The cart is moving at a constant speed or velocity

Explanation:

Equilibrium is a state of body in which it is either at rest or moves with uniform velocity. The sum of forces acting on such a body is always zero and the sum of all the torques acting on it is also zero.

There are two types of equilibrium as follows:

Static Equilibrium: When a body is at rest it is said to be in static equilibrium.

Dynamic Equilibrium: When a body is moving with constant velocity, then it is said to be in dynamic equilibrium.

Hence, the correct option here will be:

D. The cart is moving at a constant speed or velocity

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3 years ago

It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing

babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)

Dividing by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 = (6/5)*x² - 2*x - 24/5

Using quadratic formula

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}$

$x = \frac{2 \pm 5.2}{2.4}$

$x_1 = \frac{2 + 5.2}{2.4}$

$x_1 = 3$

$x_2 = \frac{2 - 5.2}{2.4}$

$x_2 = -1\; 1/3$

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

give an example of motion in which displacement of the particle is zero but acceleration is not zero in journey?

BartSMP [9]

Motion of a ball thrown by a person upwards and caught after some time is an example of motion in which displacement of the particle is zero but acceleration is not zero in journey.

The displacement of the ball is zero because the starting and end point of the motion are same, i.e, the person's hands.During its motion, the acceleration of ball is constant and non zero called as acceleration due to gravity, g= -9.8 m/s². The velocity of ball is continuously changing. It first decreases during the upward motion of the ball and then increases during the downward journey.The acceleration remains constant and non zero all the time.

4 0

2 years ago

Acceleration is rate of change of A-Position B-Time C-Velocity D-Speed

Drupady [299]

<h3>Answer:</h3>

[C] Velocity.

<h3>Explanation:</h3>

As we know that,

a = v - u/t

where, a = acceleration, v = final velocity, u = initial velocity and t = time taken to travel.

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2 years ago

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