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DIA [1.3K]
3 years ago
11

Select the correct answer.

Physics
1 answer:
Otrada [13]3 years ago
7 0

Answer:

c

Explanation:

this is because the car is only moving in the horizontal line ..it is moving with constant velocity ..

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What is the starting and final energy for a battery?
EleoNora [17]
Electrical energy is the starting and final energy of a battery
8 0
3 years ago
Suppose a rollerblade racer finnished a 132 meter race in 18 seconds. What is the average speed of the rollerblade racer
levacccp [35]

Answer:

As = 7.33 [m/s]

Explanation:

To solve this problem we must use the definition of the average speed, which is equal to the relationship between the distance over time.

As = x/t

where:

As = average speed [m/s]

x = distance = 132 [m]

t = time = 18 [s]

As = 132/18

As = 7.33 [m/s]

4 0
3 years ago
A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p
agasfer [191]

Answer:

\Delta P = 1961.4\,Pa

Explanation:

The difference of pressure is given by gauge pressure:

\Delta P = \rho_{w}\cdot g \cdot \Delta h

\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)

\Delta P = 1961.4\,Pa

8 0
4 years ago
Read 2 more answers
"if a 20 and a 40 ohm resistor are wired in parallel to a 20v source, what will be the current in the 40 ohm resistor?"
choli [55]

In parallel, neither resistor knows about the other one, and they have no effect on each other.

Each resistor is connected straight to the battery terminals. If the battery is hefty enough, it supplies current to each resistor as if the other one weren't there.

Current through a resistor = (voltage across it) / (its resistance).

Current = (20 V) / (40 ohms)

Current = 1/2 Ampere.

4 0
3 years ago
The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are
r-ruslan [8.4K]

Answer with explanation:

Part a)

v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s

Part b)

a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}

8 0
3 years ago
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