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2 years ago

15

Maria drove to the store, did some shopping, and then came home. During Maria's trip, when was her displacement equal to zero? A

) 0 hours B) It was never zero. C) at 0 hours and 2 hours D) from .75 to 1.25 hours

2 answers:

Lapatulllka [165]2 years ago

8 0

C - displacement from home being 0 is true whenever she’s home. Meaning she was home at the beginning of her trip & when she got back... home

Aleks [24]2 years ago

7 0

I think it’s a) but I’m not sure.

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In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

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

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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1 year ago

A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while

rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

$P_1V_1 =P_2 V_2$

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$P_1$ and $V_1$ are the initial pressure and volume values

$P_2$ and $V_2$ are the final pressure volume values

The final pressure here is the atmosphere, then

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1 year ago

At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m

Margaret [11]

Answer:

Part a)

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

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

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

Part b)

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

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

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so the distance from the edge of the pool is given as

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$L = 4.86 m$

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2 years ago

What is the maximum negative displacement a dog could have if it started its motion at +3 m?

raketka [301]

Answer:

- 3 meter

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A dog has started motion from +3 meter. ...(Given)

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Maximum negative distance = (-) (3)

Maximum negative distance = -3 meter

Hence, maximum negative distance is -3 meter.

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marysya [2.9K]

Answer:

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1 year ago