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3 years ago

6

A 30 kg block slides down a 20° ramp with an acceleration of 1.2 m/s2. What is the coefficient of kinetic friction between the b

lock and the ramp?

1 answer:

Vadim26 [7]3 years ago

5 0

24 is the answer because it’s is

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A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of

wlad13 [49]

The car's average acceleration would be 1.25m/s^2 or 1.25meters/second/second. That looks to be the fourth one you've listed.

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3 years ago

Read 2 more answers

- Explain how the Earth, the sun, and time are connected,

Radda [10]

Answer:

the rotation of earth is determining what part of the sun faces what part of earth making time the part thats away from the sun would be night and the one facing the sun itself would be day

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2 years ago

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A car, traveling at , encounters a dip in the road. The radius of curvature at the bottom of the dip is . Each of the car’s four

labwork [276]

Answer:

spring deflection is x = (v2 / R + g) m / 4

Explanation:

We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration

Let's write the equations on the Y axis of this description

Fe - W = m $a_{c}$

Where Fe is elastic force, W the weight and $a_{c}$ the centripetal acceleration. The elastic force equation is

Fe = - k x

4 (k x) - mg = m v² / R

The four is because there are four springs, R is theradio of dip

We can calculate the deflection (x) of the springs

x = (m v2 / R + mg) / 4

x = (v2 / R + g) m / 4

5 0

3 years ago

And 8 kg bowling ball is rolling along the frictionless alley

VLD [36.1K]

It will stop eventually

8 0

3 years ago

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

3 years ago