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butalik [34]
3 years ago
15

A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s.

Physics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

(a) 2400N

(b) Yes

Explanation:

(a) Mass (m) = 1.5kg, initial speed (v) = 4m/s, time (t) = 2.5milliseconds = 2.5/1000 = 0.0025seconds

F = mv/t = (1.5×4)/0.0025 = 6/0.0025 = 2400N

(b) The force would be different because the mass of the two hands and forearms is 3kg (2 × 1.5kg = 3kg)

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7 0
4 years ago
How does pressure depend upon force applied?​
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Answer:

Hey mate here's your answer ⤵️

Since pressure depends only on the force acting perpendicular to the surface upon which it is applied, only the force component perpendicular to the surface contributes to the pressure exerted by that force on that surface.

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3 years ago
A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What
andre [41]

Answer: The average velocity is 150 km/h

Explanation: 70+80=150

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3 years ago
Decelerating cor, driver having taken his foot off the accelerator and applying <br>brakes​
Dmitry [639]

Explanation:

In a manual transmission there are three pedals. From left to right, they are: Clutch, Brake, Gas. The clutch is the only pedal you press with your left foot. The other pedals – brake and gas – operate just like they do in an automatic transmission.

6 0
3 years ago
A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc
____ [38]

Answer:

Part a)

a = 0.36 m/s^2

Part b)

a = -1.29 m/s^2

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

F_x = Fcos25

F_x = 80 cos25 = 72.5 N

F_y = Fsin25

F_y = 80 sin25 = 33.8 N

now in vertical direction we can use force balance for the box to find the normal force on it

F_n + F_y = mg

F_n = (25)(9.81) - 33.8

F_n = 211.45 N

now kinetic friction on the box opposite to applied force due to rough floor is given as

F_k = \mu F_n

F_k = (0.300)(211.45) = 63.44 N

now the net force on the box in forward direction is given as

F_{net} = F_x - F_k

F_{net} = 72.5 - 63.44 = 9.065 N

now the acceleration of the box is given as

a = \frac{F_{net}}{m}

a = \frac{9.065}{25} = 0.36 m/s^2

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

F_x = 72.5 N

F_y = 33.8 N

now force balance perpendicular to inclined plane is given as

F_n + F_y = mgcos\theta

F_n = (25)(9.81)cos10 - 33.8 = 207.7 N

now the friction force opposite to the motion on the box is given as

f_k = \mu F_n

f_k = (0.300)(207.7) = 62.3 N

now the net pulling force along the inclined plane is given as

F_{net} = F_x - F_k - mgsin10

F_{net} = 72.5 - 62.3 - (25)(9.81)sin10

F_{net} = -32.38 N

now the box will decelerate and it is given as

a = \frac{F_{net}}{m}

a = \frac{-32.38}{25} = -1.29 m/s^2

7 0
4 years ago
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