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Flura [38]
3 years ago
5

An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for

ce of 0 N, and increasing at a uniform rate until the magnitude of the force is 38 N after the force has acted for 10 s. The body has a mass of 15 kg. What is the speed of the object at 10 s?
Physics
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer:

21741 s i believe

Explanation:

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The volume of a gas can be converted to moles by
BaLLatris [955]
B. Because I say so
6 0
3 years ago
What is the voltage across each resistor?
kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30  = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55  = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15  = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0
3 years ago
A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w
Zigmanuir [339]
Radial acceleration is given by

a_{rad}= \frac{v^2}{r}
where 

v=r \omega
then

a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2

Now

70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

Using the relation

\omega=2 \pi f

2 \pi f= \sqrt{ \frac{3.9}{70} }\\  \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz

Putting into rpm

\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm

8 0
4 years ago
A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
Write a question you would like to see on a quiz to calculate NET FORCE.
Pie

Answer:

How do you find the net force?

Explanation:

i think this is want you asked for?

5 0
3 years ago
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