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Zanzabum
3 years ago
11

Select the correct answer.

Physics
1 answer:
inn [45]3 years ago
3 0
I also need help pls now
You might be interested in
a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve
Ber [7]
Fc=mv^2/r so we get 

2000kg*(25m/s)^2/(80m)= 15625N of force 

hope this helps! Thank You!!

4 0
3 years ago
I need help with these questions ☺️☺️
Mariana [72]

Answer:

1. well the value may vary because of different reactions to the technology because its new to a lot of people but to younger ones its something normal and something we cant live without but with older ones its something that they have lived without for most of there life.

Explanation:

6 0
3 years ago
Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.
bazaltina [42]

Answer:

any object that has density more than 1.4

Explanation:

The object that has density more than 1.4 is denser than the honey

6 0
4 years ago
Read 2 more answers
If the mass of both objects are doubled while the
lutik1710 [3]

Answer:

The force becomes 16 times what it is now.

Explanation:

The formula for gravitational force is

F = G * m1 * m2 / r^2

When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.

If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)

But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2

Because this is in the denominator, the 1/4 is going to flip to the numerator.

So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.

Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.

3 0
3 years ago
An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of
slega [8]

Answer:

T_{1}=94.9^{o}C

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

T_{1}=T_{2}+\frac{q}{kS}

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\

Substitute the given values

S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m

The temperature of heater is then:

T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           T_{1}=94.9^{o}C

5 0
3 years ago
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