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3 years ago

The earth exerts a gravitational force of 500 N on Amy. What is Amy’s mass in kg?

Physics

1 answer:

White raven [17]3 years ago

6 0

Gravitational force of earth is given as weight of object

so here we know that gravitational force on Amy is 500 N

so we will have

$mg = 500 N$

now we will have

$g = 9.8 m/s^2$

now from above equation we will have

$m (9.8 m/s^2) = 500 N$

$m = \frac{500}{9.8} kg$

$m = 51 kg$

so the mass of the Amy is 51 kg

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Give two examples of energy being transformed from one type to another

GaryK [48]

Answer:

Explanation:

For example, an ice cube has heat energy and so does a glass of lemonade. If you put the ice in the lemonade, the lemonade (which is warmer) will transfer some of its heat energy to the ice.

8 0

3 years ago

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m

nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

Speed of the truck, u = 80 km/h
Distance, d = 32 km
Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

$(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h$

$94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min$

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0

2 years ago

noname [10]

Answer:

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Explanation:

Ptolemy believed in geocentric model of the solar system -the earth at the center and the sun, other planets and stars revolved about it. His belief was based on his observation that sun and other objects in sky seemed to revolve about the Earth as they rose and set in one day.

Copernicus while reading a book, constructed the theory of heliocentric system which was later proved through observations.

6 0

3 years ago

The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

5 0

3 years ago

A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object

Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

$Em_{f}$ = U = m g h

Initial mechanical energy, in the lower part of the track

Em₀ = K = ½ m v²

Em= $Em_{f}$

½ m v² = m g h

v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v

Initial before the crash

p₀ = M v₁₀ + 0

Final after the crash

$p_{f}$ = M v1f + m v

p₀ = $p_{f}$

M v₁₀ = M $v_{1f}$ + m v

As the shock is elastic the kinetic energy is conserved

K₀ = $K_{f}$

½ M v₁₀² = ½ M $v_{1f}$ ² + ½ m v²

Let's write the system of equations

M v₁₀ = M $v_{1f}$ + m v

M v1₁₀² = M $v_{1f}$ ² + m v²

We cleared v1f in the first we replaced in the second

$v_{1f}$ = (M v₁₀ - mv) / M

M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

2 m v₁₀ - v (m + 1) m/ M = 0

v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)

V₁₀ = 7.668 (2.68) / 10.82

v₁₀ = 1.90 m / s

5 0

3 years ago