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NARA [144]
3 years ago
9

A 25 kg box on a horizontal frictionless surface is moving to the right at a speed of 5.0 m s. The box hits and remains attached

to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.60 m. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero.
Physics
1 answer:
seropon [69]3 years ago
6 0

Answer:

Explanation:

Work done by the spring is negative .

Work done by force F creating displacement  d is given by the following expression .

Work = F x d

Both force and displacement are vector quantity .

When direction of force and direction of displacement is same , work is positive . When direction of force and direction of displacement is opposite , work is negative .

When spring is compressed , it exerts a restoring or opposing force in a direction opposite to the direction of displacement of box . Hence here force is opposite to displacement . Restoring force acts opposite to displacement . Hence work done by spring on box is negative .

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evablogger [386]

I don't have research to back it up, but exercise is working out. Fitness is taking care of your body, that includes working out and things like eating right.

8 0
3 years ago
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

6 0
3 years ago
To test the performance of its tires, a car
Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

\mu mg=m\frac{v^2}{r}

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

\mu is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

g=9.8 m/s^2

And re-arranging the equation for \mu, we can find the coefficient of static friction:

\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

5 0
3 years ago
Help pls thx!!!!!!! <br> this is my last question i hope!
irakobra [83]

Answer:

Left

Explanation:

newtons is a measure of force. Since there is more newtons(force) pushing to the left, the object will move left. the 15 newtons cancel each other out, leaving only 5 newtons pushing to one side with no resistance.

4 0
2 years ago
Read 2 more answers
A gas is placed in a storage tank at a pressure of 49.2 atm at 39.0C . As a safety device, a small metal plug in the tank is mad
Amiraneli [1.4K]

Answer:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=49.2 atm\\T_1=39.0^oC = 312.15 K\\P_2=?\\T_2=198^oC=471.15 K

Putting values in above equation, we get:

\frac{49.2atm }{312.15 K}=\frac{P_2}{471.15 K}\\\\P_2=74.26 atm

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

4 0
3 years ago
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