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3 years ago

2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential

Physics

1 answer:

Citrus2011 [14]3 years ago

6 0

Answer:

1703.24J

Explanation:

Given parameters:

Mass of brick = 7.9kg

Height of building = 22m

Unknown:

Potential energy of the brick = ?

Solution:

The potential energy of a body is the energy at rest of the body. Mathematically;

P.E = mgh

m is the mass of the brick

g is the acceleration due to gravity

h is the height of the building

Insert the given parameters and solve;

P.E = 7.9 x 9.8 x 22 = 1703.24J

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a 7 kilogram cat is resting on top of a bookshelf that is 2 meters high. what is the cat’s gravitational potential energy relati

attashe74 [19]

PE = mgh
Mass, m = 7kg, g ≈ 10 m/s², height = 2m

= 7*10*2

= 140 Joules.

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3 years ago

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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

mars1129 [50]

Answer:

ω2 = 216.47 rad/s

Explanation:

given data

radius r1 = 460 mm

radius r2 = 46 mm

ω = 32k rad/s

solution

we know here that power generated by roller that is

power = T. ω ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2

ω2 = $\frac{460\times 32}{34\times 2}$

ω2 = 216.47

8 0

3 years ago

A certain car is accelerating at 5.0 km/h/s. Explain what is happening to the motion of the car.

Law Incorporation [45]

The car is moving faster/accelerating quicker

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3 years ago

A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha

cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

$k=\dfrac{4\pi ^2m}{T^2}$

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

$k=\dfrac{4 \pi \times 0.2}{(1.2)^2}$

$k=\dfrac{39.48\times 0.2}{1.44}$

$k=\dfrac{7.90}{1.44}$

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

To know more about spring constant follow

brainly.com/question/1968517

#SPJ4

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2 years ago

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

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3 years ago