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3 years ago

an object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of

1 answer:

5 0

An object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of θ with the horizontal.

1. Which direction does the force of gravity act on the object?

2. Which direction does the normal force act on the object?

Which force is responsible for the object moving down the incline?

Answer:

The answer is below

Explanation:

1. When an object is moving with a constant velocity, the direction the force of gravity act on the object is DIRECTLY DOWN.

2. When an object is moving with a constant velocity, the direction the normal force act on the object "perpendicular to the surface of the plane."

3. When an object is moving with a constant velocity, the force that is responsible for the object moving down the incline is "the component of the gravitational force parallel to the surface of the inclined plane."

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A dog is running at 15 m/s and then slows down to 10 m/s. It takes him two seconds to do this. What is the dog's acceleration

OLga [1]

Explanation:

a = (Vf-Vi)/t

= (10m/s - 15m/s)/2s

a = -5 / 2 m/s^2

• so the dog diaccelerate at the rate of 5/2 m/s^2

hope it helps

6 0

3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

dolphi86 [110]

In the first case, the force acting on the spring is the weight of the mass:
$F=mg=(2.0 kg)(9.81 m/s^2)=19.6N$
This force causes a stretching of $x=6.4 cm=0.064 m$ on the spring, so we can use these data to find the spring constant:
$k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m$

In the second case, the first mass is replaced with a second mass, whose weight is
$F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N$
And since we know the spring constant, we can calculate the new elongation of the spring:
$x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm$

5 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy