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mario62 [17]
3 years ago
15

Charlotte is driving at 66.5 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 3

.47 s. How far has Charlotte traveled in feet during this time
Physics
1 answer:
Alborosie3 years ago
7 0

Answer:

the distance traveled by Charlotte in feet is 338.44 ft

Explanation:

Given;

speed of Charlotte, u = 66.5 mi/h

time of motion, t = 3.47 s

The distance traveled by Charlotte in feet is calculated as;

Distance = Speed \ \times \ time \\\\D = ut\\\\D = (\frac{66.5 \ mi}{h} \times  \frac{5280 \ ft}{1 \ mi} \times \frac{1 \ h}{3600 \ s} )(3.47 \ s)\\\\D = 338.44 \ ft

Therefore, the distance traveled by Charlotte in feet is 338.44 ft

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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

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while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

where E = electric field intensity

5 0
3 years ago
A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?
STatiana [176]
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5 0
4 years ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
aleksklad [387]

Answer:

230 N

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T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

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Dmitrij [34]

Answer: B. 1.5 meters

Explanation:

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