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3 years ago

QUESTION 24

Physics

1 answer:

lara31 [8.8K]3 years ago

6 0

-30. 40 east on timeline, 40 down, then 10 over right

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A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a

marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$ .

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$ .

The initial velocity:

$v_0 = - 20 \frac{m}{s}$ ,

take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$ .

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$ .

$y(6 \ s) = \ 1000 m \ - 300 m$ .

$y(6 \ s) = \ 700 m$ .

So this its the height of the ball 6 seconds after being thrown.

6 0

3 years ago

A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c

Reika [66]

When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ

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3 years ago

Two insulated current-carrying wires (wire 1 and wire 2) are bound together with wire ties to form a two-wire unit. The wires ar

AysviL [449]

Answer:

I'm not sure, My best friend knows, but I don't...

Explanation:

I'm sorry :)

7 0

3 years ago

During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in

andreyandreev [35.5K]

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

7 0

3 years ago

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b

Romashka-Z-Leto [24]

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

4 0

3 years ago