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STatiana [176]
3 years ago
15

A baby stroller is a rest on top of a hill which is 10 m high. The stroller and baby have a mass of 20 kg. What is the potential

energy of the baby and stroller?
Formula: PE = m g h (Gravity near earth is 9.8 m/s²)
Physics
1 answer:
Anettt [7]3 years ago
6 0

The potential energy is defined as the energy is contained in the body due to the height over the surface of the earth and it is calculated from the equation

PE=mgh

<em>where:</em>

  • PE: the potential energy in Joules.
  • m: the mass of the body in kg.
  • g: the acceleration due to the gravity in m/s^2
  • h: the height of the body over the earth in meters.

<em>in our problem:</em>

PE=20*9.8*10=1960 J

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A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an
ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02

Read more on waves here

brainly.com/question/25699025

#SPJ4

complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

7 0
2 years ago
If y^2= 3.249 x 10^-11, y = ?
Leni [432]

Answer:

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

5 0
3 years ago
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The image of an object formed by plane mirror is​
m_a_m_a [10]
Answer of your question is in this photo

8 0
2 years ago
Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

t=102.8 minutes

Explanation:

Given: m=500kg, r=1.0m, w=200\pi rad/s

a). The rotational kinetic energy

K_R=\frac{1}{2}*I*w^2

I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2

K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s

t=6168.5s\frac{1minute}{60s}=102.8 minutes

3 0
3 years ago
Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on
mrs_skeptik [129]

Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

dU = -2.59x10^4 - 6.46^4

dU = -9.05x10^4 kJ

8 0
3 years ago
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