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3 years ago

8

Jessica jogs on a path that is 25 km long to get to the park that is south of the jogging path if it takes just a cut 2.5 hours

then what is her speed and what is her velocity

1 answer:

Vinil7 [7]3 years ago

8 0

Answer:

10km/h

Explanation:

25÷2.5=10

speed = distance/time

velocity is a magnitude and needs a direction which wasnt provided so its basically the same as the speed

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What will happen if more solute is added than can be dissolved .

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A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou

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The potential energy of the lemming is 1.53 J

Explanation:

The potential energy (PE) of an object is the energy possessed by the object due to its position in the Earth's gravitational field, and it is given by:

$PE=mgh$

where:

m is the mass of the object

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h is the height of the object relative to the ground

In this problem:

m = 0.0780 kg is the mass of the lemming

We want to find the potential energy when the height is

h = 2.00 m

Therefore, we find:

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3 years ago

What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?

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Answer:

100,048

Explanation:

K.E = 1/2 m (v)^2

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3 years ago

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

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Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

3 years ago