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enyata [817]
3 years ago
9

What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?

Physics
1 answer:
NeX [460]3 years ago
6 0
The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
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The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet
algol13

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

g = \frac{Gm}{r^2}

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2 --------------------- equation (1)

FOR PLANET 2:

g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\

using equation (1):

g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}

<u>g₂ = 11 m/s²</u>

8 0
2 years ago
A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-
Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0
3 years ago
Julian is having trouble coping with stress and setting goals. What is he most likely struggling with?
klio [65]
I would say Emotional health. Just because it is more sensable
6 0
3 years ago
Read 2 more answers
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
2 years ago
PLSS HELP ITS TIMED <br> WILL MARK BRAINLIEST
dlinn [17]

Answer:

tjfd

Explanation:

tuff

4 0
3 years ago
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