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2 years ago

Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

Physics

1 answer:

liubo4ka [24]2 years ago

7 0

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

$a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N$

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

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Read 2 more answers

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog $\left ( m_d\right )=10kg$

mass of boat $\left ( m_b\right )=46kg$

distance moved by dog relative to ground= $x_d$

distance moved by boat relative to ground= $x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0= $m_d\times x_d+m_b\dot x_b$

0= $10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting $x_d$ value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

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4 0

2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i

sammy [17]

Answer:

$-0.25 rad/s^2$

Explanation:

The equivalent of Newton's second law for rotational motions is:

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = -12.5 Nm$ (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

$I=50.0 kg m^2$ is the moment of inertia

Solving for $\alpha$ , we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2$

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2 years ago

A 55.0-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high. What is the speed of the ball after it

PolarNik [594]

Answer:

The speed of the ball is 9.07 m/s.

Explanation:

Given that,

Mass of the lead ball, m = 55 kg

Height of the tower, h = 55 m

We need to find the speed of the ball it has traveled 4.20 m downward, x = 4.2 meters

The initial speed of the ball will be 0 as it was at rest initially. Let v is the speed of the ball after it has traveled 4.20 m downward. It is a case of equation of motion such that :

$v^2-u^2=2ax$

$v^2=2ax$

Here, a = g

$v^2=2\times 9.8\times 4.2$

v = 9.07 m/s

So, the speed of the ball is 9.07 m/s. Therefore, this is the required solution of given condition.

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2 years ago