Answer:
The length of the vector.
Explanation:
Forces are typically represented by vectors, which is an arrow in laymen’s terms. Vectors have a direction and a magnitude. The magnitude (or greatness of it) is shown by the length. The arrow represents the direction.
Answer:
the hypothalamus part of your brain sends a message to your body telling it to sweat.
When your internal temperature rises, your sweat glands release water to the surface of your skin.
Answer:
A= 148.92 m/s²
Explanation:
Given that
U(x,y) = (6.00 )x² - (3.75 )y ³
m= 0.04 kg
Now force in the x-direction
Fx= - dU/dx
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dx= 12 x
When x=0.4 m
dU/dx= 12 x 0.4 = 4.8
So we can say that
Fx= - 4.8 N
From Newtons law
F= m a
- 4.8 = 0.04 x a
a = -120 m/s²
Acceleration in x direction ,a = -120 m/s²
In y -direction
F= - dU/dy
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dy = 0 - 3.75 x 3 y²
When y = 0.56 m
dU/dy = - 3.75 x 3 x 0.56 x 0.56
dU/dy = - 3.52
So we can say that force in y -direction
F= 3.52 N
F= m a'
3.52 = 0.04 x a'
a'=88.2 m/s²
acceleration in y direction is 88.2 m/s²
The resultant acceleration


A= 148.92 m/s²
Answer:

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♨ Question :
- A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?
♨ 
☄ Given :
- Initial velocity ( u ) = 0
- Final velocity ( v ) = 60.0 m/s
- Time ( t ) = 7.50 s
☄ To find :
✒ We know ,

Substitute the values and solve for a.
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➛ 
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✑ Additional Info :
- When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
- When a moving object comes in rest , in the case , final velocity ( v ) = 0
- If the object is moving with uniform velocity , in the case , u = v.
- If any object is thrown vertically upwards in the case , a = -g
- When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.
Hope I helped!
Have a wonderful time ツ
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Answer:
Explanation:
a. Landing height is
H=1.3m
Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft
u=1.3m/s
Velocity of lander at impact, i.e final velocity is needed
v=?
The acceleration due to gravity is 0.4 times that of the one on earth,
Then, g on earth is approximately 9.81m/s²
Then, g on Mars is
g=0.4×9.81=3.924m/s²
Then using equation of motion for a free fall body
v²=u²+2gH
v²=1.3²+2×3.924×1.3
v²=1.69+10.2024
v²=11.8924
v=√11.8924
v=3.45m/s
The impact velocity of the spacecraft is 3.45m/s
b. For a lunar module, the safe velocity landing is 3m/s
v=3m/s.
Given that the initial velocity is 1.2m/s²
We already know acceleration due to gravity on Mars is g=3.924m/s²
The we need to know the maximum height to have a safe velocity of 3m/s
Then using equation of motion
v²=u²+2gH
3²=1.2²+2×3.924H
9=1.44+7.848H
9-1.44=7.848H
7.56=7.848H
H=7.56/7.848
H=0.963m
The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m