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Papessa [141]
3 years ago
13

a 9.0-kg crate rests on a frictionless surface. it then experiences a net horizontal force of 45. newtons. what is the resulting

acceleration of the crate?
Physics
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

5m/s^{2} .

Explanation:

Mass of the crate (m) = 9.0 Kg

Horizontal force applied to the crate(F) = 45 N .

Now,

According to the Newton's Law's of motion ,

F = ma , where a represents the acceleration of body of mass m when subjected to the horizontal force F under frictionless conditions .

Thus,

45 = 9a

a = 5 m/s^{2} .

Thus , the resulting acceleration of the crate will be 5m/s^{2} .

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Suppose a rollerblade racer finnished a 132 meter race in 18 seconds. What is the average speed of the rollerblade racer
levacccp [35]

Answer:

As = 7.33 [m/s]

Explanation:

To solve this problem we must use the definition of the average speed, which is equal to the relationship between the distance over time.

As = x/t

where:

As = average speed [m/s]

x = distance = 132 [m]

t = time = 18 [s]

As = 132/18

As = 7.33 [m/s]

4 0
3 years ago
Charged particles q1=− 4.80 nC and q2=+ 4.80 nC are separated by distance 3.00 mm , forming an electric dipole. The charges are
Dafna1 [17]

Answer:

Electric field, E = 936.19 N/C

Explanation:

It is given that,

Charge 1, q_1=-4.8\ nC=-4.8\times 10^{-9}\ C

Charge 2, q_2=+4.8\ nC=+4.8\times 10^{-9}\ C

Distance between them, d = 3 mm = 0.003 m

Torque, \tau=8\times 10^{-9}\ N-m

Angle between electric field and line connecting the charge, \theta=36.4^{\circ}

We need to find the torque exerted on the dipole. The torque experienced by the dipole in the electric field is given by :

\tau=pE\ sin\theta

p is the dipole moment, p=qd

\tau=qdE\ sin\theta

E=\dfrac{\tau}{qd\ sin\theta}

E=\dfrac{8\times 10^{-9}}{4.8\times 10^{-9}\times 0.003\ sin(36.4)}

E = 936.19 N/C

So, the magnitude of electric field on the dipole is 936.19 N/C. Hence, this is the required solution.

5 0
3 years ago
A force of 25 newtons moves a box a distance of 4 meters in 5 seconds.
Svetllana [295]

Answer:

100 Nm and 25Nm/s.

Explanation:

<u>Given the following data;</u>

Force = 25N

Distance = 4m

Time = 5secs

To find the workdone;

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 25*4

Workdone = 100 Nm

To find the power consumed;

Power = workdone/time

Substituting into the equation, we have;

Power = 100/4

Power = 25Nm/s

The work done on the box is 100 Nm, and the power is 25 Nm/s.

5 0
3 years ago
A projectile is fired at an upward angle of 29.7° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its
german

Answer:

79.2 m/s

Explanation:

θ = angle at which projectile is launched = 29.7 deg

a = initial speed of launch = 130 m/s

Consider the motion along the vertical direction

v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s

y = vertical displacement = - 108 m

a = acceleration = - 9.8 m/s²

v = final speed as it strikes the ground

Using the kinematics equation

v² = v₀² + 2 a y

v² = 64.4² + 2 (-9.8) (-108)

v = 79.2 m/s

5 0
3 years ago
Course Task
kaheart [24]

Answer:

gravity effects only the mass of an object

Explanation:

an acorn is heavier that a stick and it's round shape gives it less wind resistance so gravity pulls it down faster. A stick vertical has little wind resistance but when rotated to when it is horizontal it has more resistance so it slows down but the the force of gravity is still as strong. A leaf has very little mass and it's shape provideds lots of wind resistance so it falls slower. The moon is not only effected by the earth's gravity it is also effected by the suns. It has its own gravity as well. with all these factors it stays in place and moves with the earth.

4 0
3 years ago
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