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faltersainse [42]
3 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]3 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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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

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2 years ago
elements found in the same ______ have similar chemical properties. What should go in the blank? Row or column?
Temka [501]
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suppose the same chest is at rest. you push it horizontally with force of 110N but it does not budge. What is the contact force
Radda [10]

Answer:

110 N

Explanation:

When a force is applied on a body and body does not move, it means the body remains at rest.

In this condition, there is a contact force between the body and the floor which is called static friction.

Th static friction force is a self adjusting force and comes into play when the body is at rest.

Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.

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