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A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How f

kiruha [24]

Work is obtained by multiplying the force and the object's displacement. The force and displacement and force should be in the same direction in order to have work.
W = F x d
d = W / F
Substituting the known values,
d = 352 J / 45 N = 7.82 m
Thus, the displacement of the student is 7.82 m.

8 0

4 years ago

TRUE OR FALSE The time period of a simple pendulum does not depend upon mass, but depends, but depends on the size of the bob, b

Dimas [21]

Time period depends on the length of the pendulum and the force of gravity.

<h3>On what things time period depends?</h3>

The time period of a simple pendulum does not depend on mass and material of the bob. It depends on the length of the pendulum and the force of gravity because more length of the pendulum leads to more time to complete one cycle.

So we can conclude that time period depends on the length of the pendulum and the force of gravity.

Learn more about time here: brainly.com/question/2854969

#SPJ1

3 0

2 years ago

Match the theory to the statement that best describes it. 1. big bang. 2. steady state. 3.osscillating universe. 4. inflation Ch

Gre4nikov [31]

Explanation :

(1) Big bang : (1) The most accepted theory on the origin of the universe.

This theory shows the expanding of the universe from high density and high-temperature states.

(2) Steady state : (3) All is the same and will always stay the same.

Steady state means that the properties of any system remain the same always.

(3) Oscillating universe : (4) Agrees with the big bang theory but insists the universe expanded much quicker.

Oscillating universe theory is the result of big bang theory.

(4) Inflation Choices : (2) it's like an inflating and deflating balloon that never stops.

In cosmology, cosmic inflation or deflation is just the expanding and contraction of the universe.

So, the statements and the choices are related as:

(1)-(1)

(2)- (3)

(3)-(4)

(4)-(2)

8 0

4 years ago

Read 2 more answers

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

k0ka [10]

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago

7 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

4 years ago