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1 year ago

11

A friend climbs an apple tree and drops a 0.22-kg apple from rest to you, standing 3.5 m below. When you catch the apple, you br

ing it to rest in 0.28 s.
A.) What was the speed of the apple just before you caught it?

B.) What average force did you exert on the apple to bring it to rest? (Be sure to include both the weight of the apple and the force needed to bring it to rest.)

1 answer:

ella [17]1 year ago

3 0

Answer: (A) velocity = 2.8m/s

(B) Average force = 1.9536 Newtons

Explanation:

Find detailed explanation in the attached picture

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vector A makes equal angles with x,y and z axis. value of its components (in terms of magnitude of vector A will be?

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X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared

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1 year ago

Read 2 more answers

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

$mv_o=m(v_1 + v_2)$ where m is the mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are velocity of the impacting object after collision and velocity after impact of the originally constant object

$5m=m(v_1 +v_2)$

Therefore $v_1+v_2=5$

After collision, kinetic energy doubles hence

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting 5 m/s for $v_o$ then

$2*(5^{2})= v_1^{2} + v_2^{2}$

$50= v_1^{2} + v_2^{2}$

Also, it’s known that $v_1+v_2=5$ hence $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then $v_2=6.83 m/s$

Substituting, $v_1=-1.83 m/s$

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0

1 year ago

If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?

Sav [38]

<h2>Complete Question:</h2>

You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.60 kΩ .

(a) If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?

(b) How much electrical power is delivered to your body?

<h2>Answer:</h2>

(a) 0.075A

(b) 9W

<h2>Explanation:</h2>

The voltage (V) passing across or supplied to a body is directly proportional to the current (I) flowing through the body as stated by Ohm's law. i.e

V ∝ I

=> V = I x R ----------------------(i)

Where;

R = constant of proportionality called resistance of the body

(a) As stated in the question;

The body is wet and thus will conduct electricity and has the following;

V = voltage supplied = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substitute these values into equation(i) as follows;

120 = I x 1600

Solve for I;

I = $\frac{120}{1600}$

I = 0.075A

Therefore the amount of current that will pass through your body is 0.075A

(b) Electrical power(P), which is commonly measured in Watts(W), delivered to a body is the product of the current(I) and voltage (V) supplied to the body. i.e

P = I x V ---------------------(ii)

Where;

I = 0.075A [as calculated above]

V = 120V [given in the question]

Substitute these values into equation (ii) as follows;

P = 0.075 x 120

P = 9W

Therefore, the electric power delivered to your body is 9W

7 0

1 year ago

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

Korvikt [17]

Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

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1 year ago

The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at

elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

a) The point at which particle stop, it's velocity = 0 m/s

So dx/dt = 0

0 = 0- 12t = -12t

So when time t= 0, velocity = 0 m/s

So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

$x=4.00 - 6.00*0^2 = 4m$

SO at x = 4m the particle is (momentarily) stop

c) We have $x=4.00 - 6.00t^2$

At origin x = 0

Substituting

$0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}$

t = 0.816 seconds or t = - 0.816 seconds

So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

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2 years ago