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alexandr1967 [171]
3 years ago
9

Suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of the object

varies inversely with its mass. a certain force acting upon an object with mass 3 kg produces an acceleration of 15 /ms2 . if the same force acts upon another object whose mass is 5 kg , what would this object's acceleration be?
Physics
1 answer:
amm18123 years ago
6 0
Solve for "x"
X = speed of the 5 kg 
3/15=5/x
cross multiply 
3x=75 
because (15)(5)=(3)(x)
divide 3 in both side
X=25 
or 
A=25 m/s^2 
Hope this helps

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crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

F_{net} = F_1 - F_2

F_{net} = 60 - 45

F_{net} = 15 N

so boat will have net force F = 15 N in forward direction due to both jumps

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When a surfer rides an ocean wave on her surfboard, she is actually riding on. A. a crest that is toppling over. . B. a trough o
ruslelena [56]
The right answer to this question is A. a crest that is toppling over. When a surfer rides an ocean wave on her surfboard, she is actually riding on a crest. The crest is the point on a wave with the maximum value or upward displacement within a cycle.
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3 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t
kozerog [31]

Answer:

a) 6 times farther.  b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )

If the total displacement in the vertical direction is 0 (which means  that the time if the total time of flight), we can solve for t, as follows:

t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s

On earth, this time could be calculated in the same way:

t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0
3 years ago
Hey pls help thanks a lot
galben [10]

Answer:

I am not sure about the answer as I don't have a proper calculator besides me now

Explanation:

but I used this equation:

(8.20)sin30(1-d)=10d

Idk whether it is correct or not, I'm just a student too

what is your method of doing this question?

4 0
3 years ago
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