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2 years ago

More damage is done and more life is endangered in a head-on collision of two cars, each traveling at 30 mph, than in a

Physics

1 answer:

seropon [69]2 years ago

6 0

Answer:

More damage is done and more life is endangered in a head-on collision of two cars, each traveling at 30 mph, than in a car crashing into a brick wall at 30 mph because the relative velocities of the head-on collision are the sum of the two velocities.

Explanation:

Car crashing to a brick wall at 30 mph is less dangerous than two cars travelling in opposite direction at 30 mph each.

This is because, for one car, the other is at a relative speed of 30+30 = 60 mph.

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An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with

a_sh-v [17]

Explanation:

We assume that length of boat is L cm. And, we suppose to have a coordinate system relative to the bottom of the lake with:

Boat's position initially from x = 0 cm to x = L cm

Let person (S) is initially at x = L cm (right hand end of boat)

and, let boat's center of mass at x = C cm

Therefore, initially center of mass of overall (system's) is at $X_{1}$ where (taking moments about x = 0).

$(80 + 130)X_{1}$

= 130C + 80L ........... (i)

After S walks left to the other end of the boat, the boat moves right 80 cm.

Hence,

Boat's position is now from x = 80 cm to x = (L+ 80) cm

S's position is x = 80 (left end of boat)

Boat's center of mass is at (C + 80).

Therefore, overall (system's) final center of mass is at $X_{2}$ where (taking moments about x = 0).

$(80 + 130)X_{2}$

= $(130)(C + 80) + 80 \times 80$ ............. (ii)

As no external force is acted on the system, the system's center of mass has not moved, so $X_{1} = X_{2}$ .

This means the left hand sides of both equations (i) and (ii) are equal. So, we can equate the right hand sides giving: the equation as follows.

130C + 80L = $(130)(C+80) + 80 \times 80
$

80L = 16800 cm

L = 210 cm

= 2.1 m

Length of boat = 2.1 m

When the person is walking from one end to the other as there was no external force then center of mass of the person-boat system did not move.

6 0

2 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

2 years ago

When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is

Maurinko [17]

Answer:

Part a)

$\omega = 33.33 rad/s$

Part b)

$F = 500 N$

Part c)

$R_{max} = 40.8 m$

Explanation:

Part a)

As we know that the velocity of the tip of the kicker's shoe is given as

$v = 35 m/s$

also the length of the tip of the shoe from his hip joint is given as

$L = 1.05 m$

now the angular speed is given as

$\omega = \frac{v}{L}$

$\omega = \frac{35}{1.05}$

$\omega = 33.33 rad/s$

Part b)

As we know that force on the ball is given as rate of change in momentum of the ball

so it is given as

$F = \frac{\Delta P}{\Delta t}$

so we have

$F = \frac{m(v_f - v_i)}{\Delta t}$

$F = \frac{0.500(20 - 0)}{20 \times 10^{-3}}$

$F = 500 N$

Part c)

As we know that the formula of range is given as

$R = \frac{v^2 sin(2\theta)}{g}$

now for maximum range we know

$\theta = 45 degree$

$R_{max} = \frac{v^2}{g}$

$R_{max} = \frac{20^2}{9.8}$

$R_{max} = 40.8 m$

6 0

2 years ago

Elyse is explaining to a friend that she will be conducting a scientific investigation. Her question is "How many leatherback se

trapecia [35]

The scientific investigation has five basic steps and one feedback step: (1) to make an observation (2) to ask a question (3) hypothesize (4) predict and (5) make a new hypothesis based on the results. What is unusual about the description of Elyse’s scientific investigation is she did not follow the basic steps. She should have done observation first before asking a question and forming a hypothesis.

8 0

2 years ago

The diagram shows the electric field due to point charge Q. Which statements are correct? Check all that apply.

Bess [88]

Answer:

The correct statements are:

A. The electric field is nonuniform.

D. Charge Q is positive.

E. If charge A moves toward charge Q, it must be a negative charge

Explanation:

The answer choices are:

A. The electric field is nonuniform.

B. The electric field is uniform.

C. Charge Q is negative.

D. Charge Q is positive.

E. If charge A moves toward charge Q, it must be a negative charge.

F. If charge A moves toward charge Q, it must be a positive charge.

<h2>Solution</h2>

The electric field is the electrostatic force per unit of charge,

$\vec E=\dfrac{\vec F}{Q}$

around around a charge, where another charge would experience the electrostatic force.

The electric field lines are shown in a diagram with arrows ditributed radially away from a positive charge and radially toward a negative charge.

Since the arrows are away from Q, Q is a positive charge: statement D.

Since the size of the arrows decreases as you move away from Q the stregth of the field is not uniform: statement A.

Since the charge Q is positive, a negative charge would be attracted toward it: statement E.

4 0

2 years ago