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2 years ago

Are predators or their prey more likely to be successful?

Physics

2 answers:

puteri [66]2 years ago

8 0

Predators are more likely to be more successful because they are typically bigger and more powerful than the prey

mihalych1998 [28]2 years ago

3 0

Predator because they are the one’s preying on the animals, the chances of the prey getting away is slim.

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Choose all the answers that apply.

Serhud [2]

According to Newton's first law of motion:

-- objects at rest will remain at rest unless acted upon by an outside force

-- objects in motion will remain in motion unless acted upon by an outside force

8 0

3 years ago

The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i

Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

$\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170$

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

4 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

3 years ago

A downward force of magnitude 5 n is exerted on the book by the force of.

WINSTONCH [101]

Answer:

A downward force of magnitude 5 N is exerted on the book by the force of ... Do the downward force in Part A (gravity) and the upward force in Part B ...

6 0

1 year ago

A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0

IgorLugansk [536]

Answer:

Magnitude of the force is

$F_3 = 2.83$