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1 year ago

What are three kinds of forces that affect an object’s motion? What effects do these forces have?

Physics

1 answer:

Harman [31]1 year ago

4 0

Answer:

Different forces (including magnetism, gravity, and friction) can affect motion

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How much work do you do when you push a shopping cart with a force of 20 N for a distance of 5m?

KiRa [710]

Answer:

A.100J

Explanation:

Given parameters:

Force on car = 20N

Distance = 5m

Unknown:

Work done = ?

Solution:

Work done is the product of force and distance;

Work done = force x distance;

Insert given parameters and solve;

Work done = 20 x 5 = 100J

6 0

2 years ago

Experiment to determine the refractive index of a glass prism FOCAL LENGTH = 9.5cm i° e° d° 30 43 69 40 41 61 45 39 56 50 37 48

yarga [219]

We are given the data for the angle of incidence and angle of refraction.
The first part of the problem is to plot the deviation and the angle of incidence. The minimum deviation is 37 with 65 degrees as the corresponding angle of incidence. The maximum deviation is 69 with a corresponding angle of incidence of 30 degrees. It is not advisable to use small values for the angle of incidence since it would result to a higher deviation from Snell's Law.

3 0

2 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

Vikki [24]

Answer:

The force constant is $k =1.316 *10^{7} \ N/m$

The energy stored in the spring is $E = 1.68 *10^{7} \ J$

Explanation:

From the question we are told that

The mass of the object is $M = 4.8*10^{5} \ kg$

The period is $T = 1.2 \ s$

The period of the spring oscillation is mathematically represented as

$T =2 \pi \sqrt{ \frac{M}{k}}$

where k is the force constant

So making k the subject

$k = \frac{4 \pi ^2 M }{T^2}$

substituting values

$k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}$

$k =1.316 *10^{7} \ N/m$

The energy stored in the spring is mathematically represented as

$E = \frac{1}{2} k x^2$

Where x is the spring displacement which is given as

$x = 1.6 \ m$

substituting values

$E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2$

$E = 1.68 *10^{7} \ J$

7 0

2 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$