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Fynjy0 [20]
3 years ago
7

1 6. Which of the following astronomical units is c/osest to the distance of the earth to the sun?

Physics
1 answer:
Ganezh [65]3 years ago
6 0

Answer:

astronomical unit

Explanation:

https://en.wikipedia.org/wiki/Astronomical_unit

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A force acts on a body of mass 13 kg initially at rest. The force
PtichkaEL [24]

Answer:

Force that acted on the body was F = 13 N

Explanation:

If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s

When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

v_f=v_i+a*t

which in our case becomes;

10\,m/s=0+a*(10\,s)

and we can solve for the acceleration as:

a = 10/10  m/s^2 = 1 m/s^2

Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:

F = 13 kg * 1 m/s^2 = 13 N

4 0
3 years ago
As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0
Alika [10]

velocity of the physics instructor with respect to bus

v = 6 m/s

acceleration of the bus is given as

a = 2 m/s^2

acceleration of instructor with respect to bus is given as

a = -2 m/s^2

now the maximum distance that instructor will move with respect to bus is given as

v_f^2 - v_i^2 = 2 a d

0 - 6^2 = 2(-2)(d)

-36 = - 4 d

d = 9 m

so the position of the instructor with respect to door is exceed by

\delta x = 9 - 6 = 3 m

so it will be moved maximum by 3 m distance

7 0
3 years ago
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w
Ann [662]

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: P*V=n*R*T

where P is the gas' pressure, V its volume, n the number of moles of gas,  R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450

So the pressure must have gone up to 450 kilopascals.

3 0
2 years ago
Is it possible to calculate the displacement based on an elapsed time from a position time graph?
Nataly [62]
No I don’t think so. But it worth a try tho. Try it out.
5 0
2 years ago
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