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borishaifa [10]

3 years ago

13

imagine that you have a rectangular swimming pool, but you have no idea how deep it is. you know that the pool has a volume of 2

16m3. you measured the dimensions of the pool and found that it is 15 m long and 6 m wide.

1 answer:

r-ruslan [8.4K]3 years ago

6 0

Answer:

To calculate the volume of any rectangular shape,

We can simply multiply length to wide and to depth.

Let d be the depth.

15 x 6 x d = 216

d = 216 / 15 / 6

d = 2.4

Therefore the depth is 2.4m

Suppose a box is 8 cm long, 6 cm wide, and 4 cm high. To calculate its volume, you would multiply

the area of the object by its height:

area = 8 cm x 6 cm

area = 48 cm2

volume = 48 cm2

x 4 cm

volume = 192 cm3

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A father racing his son has half the kinetic energy of the son, who has half the mass of the father.The father speeds up by 1.0

Helga [31]

Answer:

(a) $v_f=2.414\ m.s^{-1}$

(b) $v_s=4.828\ m.s^{-1}$

Explanation:

Let:

mass of father be, $m_f$

mass of son be, $m_s=\frac{m_f}{2}$

speed of son be, $v_s$

initial speed of father be, $v_f$

After speeding up, speed of father is $v_f+1$

<u>We know Kinetic Energy is given as</u>

$KE=\frac{1}{2} m.v^2$ .....................................(1)

where:

m = mass

v = velocity

Hence, according to the initial condition the father is having kinetic energy half the kinetic energy of the son.

$KE_s=2.KE_f$

$\frac{1}{2} m_s.v_s^2=2\times \frac{1}{2} m_f.v_f^2$

$(\frac{m_f}{2})\times v_s^2=2\times m_f\times v_f^2$

$v_s=2v_f$ .................................................(2)

According to the final condition:

$\frac{1}{2} m_s.v_s^2= \frac{1}{2} m_f.(v_f+1)^2$

$(\frac{m_f}{2})\times v_s^2=m_f.(v_f+1)^2$

$v_s^2=2(v_f+1)^2$

$v_s=\sqrt{2}(v_f+1)$ .....................................................(3)

(a)

From eq. 2 & 3

$2v_f=\sqrt{2}(v_f+1)$

$\sqrt{2}\ v_f=(v_f+1)$

$v_f(\sqrt{2}-1)=1$

$v_f=\frac{1}{(\sqrt{2}-1)}$

$v_f=2.414\ m.s^{-1}$

(b)

<em>putting the above value in eq. (2)</em>

$v_s=4.828\ m.s^{-1}$

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What do we call seismic waves that are transmitted along the outside of earth?

crimeas [40]

Surface waves fits that description.

8 0

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Describe the magnetic field in a bar magnet. Enter your answer in the spave provided.

neonofarm [45]

Answer:

The lines of the magnetic field from a bar magnet form closed lines. By convention, the field direction is taken to be outward from the North pole and into the South pole of the magnet. Permanent magnets can be made from ferromagnetic materials.

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4 years ago

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lana [24]

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8 0

3 years ago

A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s

egoroff_w [7]

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt = 10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

4 0

3 years ago