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2 years ago

Your forehead can withstand a force of about 6.0 kN before fracturing, while your cheekbone can only withstand about 1.3 kN .

Physics

1 answer:

Thepotemich [5.8K]2 years ago

8 0

Answer:

Explanation:

Maximum force for forehead = 6 kN

maximum force for cheek bone = 1.3 kN

Part A:

mass of base ball, m = 140 g = 0.14 kg

initial velocity of base ball, u = 30 m/s

final velocity of base ball, v = 0

Time, t = 0.0015 s

Acceleration is defined a the rate of change of velocity.

Use first equation of motion

v = u + at

0 = 30 + a x 0.0015

a = - 20000 m/s²

Part B:

Force = mass x acceleration

F = 0.140 x 20000

F = 2800 N

Part C:

Force applied on your head is 2800 N.

Part D:

As the force is 2800 N that means 2.8 kN which is less than 6 kN so you are safe.

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Digiron [165]

Speed is the answer

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2 years ago

An airplane is being pushed West by its engines with a force of 350 N as it starts to take off the runway at the Atlanta airport

Marizza181 [45]

Answer:

option C

Explanation:

given,

Force by the engine on plane in West direction = 350 N

Frictional force on the runway = 100 N in east

force exerted by the wind = 100 N in east

net force and direction = ?

consider west to be positive and east be negative.

when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane

Net force = 350 N - 100 N - 100 N

= 150 N

as our answer comes out to be positive so the airplane will be moving in West

hence, the correct answer is option C

8 0

2 years ago

Help me find the acceleration

ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the timeless suvat equation is likely what you need for this question.

In the timeless suvat equation,

$a = \dfrac{v^2 - u^2}{2\; x}$

where

$a$ is the acceleration of the car;
$v$ is the final velocity of the car;
$u$ is the initial velocity of the car; and
$x$ is the displacement of the car.

Note that v and u are velocities. Make sure that you include their signs in the calculation.

In this question,

$a$ is the unknown;
$v = -10.9 \; \text{m} \cdot \text{s}^{-2}$ ;
$u = -27.7 \; \text{m} \cdot \text{s}^{-2}$ ; and
$x = - 105 \; \text{m}$ .

Apply the timeless suvat equation:

$a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}$ .

The value of $a$ is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0

3 years ago

Can you answer any of these?

givi [52]

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3 0

3 years ago

Read 2 more answers

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

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3 years ago