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gogolik [260]
3 years ago
9

What is the acceleration aaa of the refrigerator 4 ss after the person begins pushing on it with a force of 400 NN

Physics
1 answer:
ruslelena [56]3 years ago
3 0

Answer:

2 ms^2

Explanation:

Given that the mass of the refrigerator  is 200Kg

The net force acting on the refrigerator is 400N

The time takes is 4s

From Newtons's second law of motion;

F =ma

a = F/m

a = 400/200

a = 2 ms^2

This is a constant acceleration

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I have a picture of the question I need help with,
eduard

Given:

The initial velocity of the object, v=30 m/s

a_t=0

a_c≠0

The time period is Δt.

To find:

The right conclusion among the given choices.

Explanation:

a_t represents the tangential accleration on the object and a_c represents the centripetal acceleration on the object.

The centripetal acceleration is the acceleration that keeps the object in its circular path. The centripetal force only changes the direction of the velocity and not the magnitude.

Thus the magnitude of the velocity of the object remains the same after a time interval of Δt. But the direction of the velocity of the object will be changed and will be unknown after Δt seconds.

Final answer:

Thus the object will be traveling at 30 m/s in some unknown direction.

Therefore, the correct answer is option a.

7 0
1 year ago
An object is moving to the west at a constant speed. three forces are exerted on the object. one force is 10 n directed due nort
SVETLANKA909090 [29]
If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
F_{12} =  \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N
with direction at 45^{\circ} north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be 45^{\circ} south-east.
6 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t
borishaifa [10]

Answer:

Angular acceleration will be 18.84rad/sec^2

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity \omega _i=0rev/sec

And final angular velocity \omega _f=24rev/sec

Time is given as t = 8 sec

From equation of motion

We know that \omega _f=\omega _i+\alpha t

24=0+\alpha \times 8

\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2

So angular acceleration will be 18.84rad/sec^2

4 0
3 years ago
When measuring accelerations with three different graphs (x-t, v-t, a-t) which gives the most reliable results and why?
kipiarov [429]

Answer:

the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

Explanation:

The different graphics depending on time give various information, let's examine what we can get from some

Graph of x -t. from this graph we can obtain the speed through the slope, but the acceleration is not directly obtainable

v-t chart. We can get the acceleration not through the slope and the distance traveled by the area under the curve. Obtaining acceleration is very accurate since it is an average that avoids possible errors in measurements. This is the best graph to find the acceleration

Graph of a-t In this graph the acceleration is a point on the Y axis, it gives some errors because it depends strongly on the possible experimental errors.

In conclusion, the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

5 0
3 years ago
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