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aleksley [76]
3 years ago
10

A 10 kg box rests on the ground. What is the weight of the box? __ N

Physics
2 answers:
Nata [24]3 years ago
5 0

Explanation:

1. Mass of the box, m = 10 kg

Weight of an object is equal to the product of its mass and acceleration due to gravity. Its expression is given by :

W = mg

W=10\ kg\times 9.8\ m/s^2

W = 98 N

2. The normal force acting on an object is equal to its weight. So, N = 98 N

3. If the box is pushed to the left with 20 N of force and it is not moving, then there is a force acting on it which is called frictional force. The static friction force have magnitude of 20 N and it is acting in opposite direction of motion.

4. When the pushing force is increased to 40 N and the box begins to move, then the maximum static frictional force is equal to applied force i.e. 40 N.

ANTONII [103]3 years ago
3 0
The weight of an object is its mass times acceleration.

Weight of the box = (10kg)(9.81m/s^2) = 98.1N

If the box is at rest, then the normal force is equal to the weight of the box:

Normal force = 98.1N

If the box does not move, then the static friction force with 20 N is also 20N

Static Friction Force = 20N

If the box begins to move at 40, the maximum static friction force is equal to 40N

Maximum static friction force = 40N
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You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.
Vikki [24]

Answer:

A) Average speed = 18.75 m/s

B) More time is spent at 15 m/s than at 25 m/s.

Explanation:

Let the first distance be d1 and the second distance be d2.

We are given;

d1 = 10 km = 10000 m

d2 = 10 km = 10000 m

Speed; v1 = 15 m/s

Speed; v2 = 25 m/s

Now, the formula for distance is; Distance = speed x time

Thus:

d1 = v1 x t1

t1 = d1/v1 = 10000/15 = 666.67 seconds

Also,

d2 = v2 x t2

t2 = d2/v2 = 10000/25 = 400 seconds

Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s

From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;

More time at 15 m/s than at 25 m/s.

5 0
3 years ago
Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it
Kay [80]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

F=kx

where, F is the force which is stretching or compressing the spring,

k is the spring constant; and

x is the distance the spring is stretched.

Substituting the given values to find the elastic constant  k to get:

F=kx

4=k(2)

k=\frac{4}{2}

k=2

Therefore, the elastic constant is 2 Newton/cm.

3 0
3 years ago
What happens to the particles of the liquid inside a thermometer when the thermometer is heated?
aev [14]
The particles of the liquid inside a thermometer speed up and spread apart when the thermometer is heated. In short, the particles expand from one another when they're heated, and become condensed and compact when chilled.

5 0
3 years ago
Read 2 more answers
Which answer explains why plate B is moving underneath plate A
blsea [12.9K]

Answer:

A.

Explanation:

Earth is composed of different layers and one layer moves over another due to differences in the densities.

According to the physics of density, a substance having less density floats over a higher density substance. The oceanic crust has more density than the continental crust that is why continental crust float over oceanic crust.

So in the given example, plate B is moving below the plate A, it means plate B is more dense than plate A because plate B is composed of oceanic crust . <u>For example : continents float over the asthenosphere (a layer below the lithosphere).</u>

Hence, the correct answer is "A ".

4 0
3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
3 years ago
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