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kenny6666 [7]
3 years ago
10

An object with a mass of 5.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the acceleration? (20 poin

ts)
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

40N

Explanation:

F=ma

F= 5 × 8 = 40N

.......

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How do you solving kinematic equations for horizontal projectiles?
daser333 [38]
See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION

7 0
3 years ago
How to solve number 10
Vinil7 [7]

The 'formulas' to use are just the definitions of 'power' and 'work':

Power = (work done) / (time to do the work)

and  

Work = (force) x (distance) .

Combine these into one. Take the definition of 'Work', and write it in place of 'work' in the definition of power.

Power = (force x distance) / (time)

From the sheet, we know the power, the distance, and the time.  So we can use this one formula to find the force.

Power = (force x distance) / (time)

Multiply each side by (time):  (Power) x (time) = (force) x (distance)

Divide each side by (distance): Force = (power x time) / (distance).

Look how neat, clean, and simple that is !

Force = (13.3 watts) x (3 seconds) / (4 meters)

Force = (13.3 x 3 / 4) (watt-seconds / meter)

Force = 39.9/4 (joules/meter)

<em>Force = 9.975 Newtons</em>

Is that awesome or what !

6 0
3 years ago
A 2-kg book falls off a shelf. It hits a student traveling 2 m/s. How much kinetic energy does the book have?
Gekata [30.6K]

4 J

1/2 mv^2 is ke formula

3 0
3 years ago
28. Sound can be heard around a corner because of
Lesechka [4]

Answer:

Diffraction of sound wavelengths.

Explanation:

Diffraction-A wave is able to bend around a corner due to the effects of diffraction. sound aves are capable of bending around corners in the same magnitude as it's wavelength making it possible to hear sounds around corners.

5 0
3 years ago
Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject
Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}

where

\mu_{sun} is gravitational parameter of sun =  1.32712 x 10^20 m^3 s^2.t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}

t = 12,105.96 sec

6 0
3 years ago
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