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ad-work [718]
3 years ago
7

A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.30, and the c

oefficient of kinetic friction is 0.20. What horizontal force must be applied to the box to cause it to start sliding along the surface
Physics
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

The horizontal force that must be applied to the box to cause it to start sliding along the surface is 162N

Explanation:

To start sliding the box on the surface it must overcome its static frictional force  under equilibrium condition

The net force on the box is

F - fs = 0

F = fs = us N = us mg

Force = ( 0.3) x ( 55 kg) x ( 9.8 m/s^2) = 161.7 approximately 162 N

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vazorg [7]

3.

a)

r = distance of each mass in each hand from center = 0.6 m

m = mass of each mass in each hand = 2 kg

v = linear speed = 1.1 m/s

L = combined angular momentum of the masses = ?

Combined angular momentum of the masses is given as

L = 2 m v r

L = 2 (2) (1.1) (0.6)

L = 2.64 kg m²/s


b)

v' = linear speed when she pulls her arms = ?

r' = distance of each mass from center after she pulls her arms = 0.15 m

Using conservation of momentum , angular momentum remains same, hence

L = 2 m v' r'

2.64 = 2 (2) (0.15) v'

v' = 4.4 m/s


4 0
4 years ago
A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.52 radians with the
vovangra [49]

Answer:

a = 8.951 m/s²

Explanation:

given,

angle = 0.52 radians

μ_s = 0.84

μ_k = 0.48

acceleration = ?

using

F +  f = m a

mg sin θ +  μk mg cos θ = m a

a = g sin θ  + μk g cos θ

a = 9.8 x  sin 0.52 + 0.48 x 9.8 x  cos 0.52

a = 4.869 + 4.082

a = 8.951 m/s²

the magnitude of acceleration is a = 8.951 m/s²

8 0
3 years ago
I need help pelase ASAP!
Mariulka [41]

1. Elements

2. Physically

3. Properties

4. Pure

5. Compounds

6. Combined

7. Original

8 0
3 years ago
A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
vredina [299]

(a) 4.0 s

The acceleration of the car is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

(c) -5.0 m/s^2

The deceleration is given by the problem, and its value is -5.0 m/s^2.

(d) 5000 N

The net force applied on the car is given by

F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0
4 years ago
Explain how bats use ultrasound to catch a prey​
Gnoma [55]

Answer:

Ultrasound is used by bats to navigate (move) and catch prey. Ultrasonic squeaks are produced by bats. These squeaks ponder prey and then return to the bat's ear. This provides bats with information about the location of prey, allowing them to catch it.

Explanation:

8 0
3 years ago
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