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faltersainse [42]
2 years ago
9

5 A photographer runs away from an angry rhino toward the safety of her Jeep What will happen? (Speeds and distances are shown b

elow the image.)
O
O
o
- Photographer is 10 meters from the Jeep
--Photographer runs 6 meters per second,
-Rhinois 16 meters from the Jeep
-Rhino runs 8 meters per second
- The Jeep is not moving
o
.
O A. The photographer will get to the Jeep before the rhino catches her
O B. The rhino will catch the photographer before she reaches the Jeep
O . The rhino and the photographer will reach the Jeep at the same time,
OD. More information is needed

Physics
1 answer:
irinina [24]2 years ago
8 0

Answer:

A. The photographer will get to the jeep before the rhinocerous

Explanation:

Δv =  Δd/Δt

we can rearrange for time

Δt = Δd/Δv

For the photographer:

Distance is 10m and moves at 6m/s

Δt = 10m/6m/s

Δt = 1.67s

For the rhinocerous

Distance is 16m and moves at 8m/s

Δt = 16m/8m/s

Δt = 2.00s

The distances were to get to the jeep, the photographer makes it to the jeep in a shorter amount of time than the rhino

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We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with
amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

6 0
2 years ago
A car is moving with a constant velocity of 25 m/s. Which of the following is true?
S_A_V [24]

b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

a) the force from the engine is greater than all the forces of friction.  --> FALSE. In fact, the car is moving at constant velocity: this means that its acceleration is zero,

a = 0

and so Newton's second law becomes

\sum F = ma = 0

where \sum F is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero.  --> TRUE, for what we said at point A)

c) the inertia is changing.  --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car.  --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

F_f = \mu mg

where \mu is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

e) All of the above. --> FALSE

Learn more about net force and Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

6 0
3 years ago
How do butterfly dest on there wings
4vir4ik [10]
They have scales and they rub off easily
6 0
3 years ago
If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice
Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

8 0
3 years ago
an auditorium measures 30.0 m ✕ 15.0 m ✕ 5.0 m. the density of air is 1.20 kg/m3. (a) what is the volume of the room in cubic fe
Fynjy0 [20]

dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

density = 1.20 kg/m3

(a)volume = lenght * breadth * height

      = 30 * 15 * 5

      = 2250 metre cube = 2.25 cubic meter

(b)   mass of air = density * volume

        mass of air = 1.2 * 2250

mass of air = 2700kg

weight  = mass * 9.8

             = 2700 * 9.8

             = 26,460 N

  • The definition of Density is the amount of matter in a given space, or volume
  • Density = mass/volume
  • units for density kg/m^3
  • Density of water 1g/ml
  • Salt water is denser that is why  don't sink as easily.

To know more about density  visit : brainly.com/question/15164682

#SPJ4

5 0
1 year ago
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