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3 years ago

14

While playing red light green light, Shelbie starts at rest and accelerates at a rate of 1.2 m/s2 constantly. Shelbie traveled 1

5 meters, how long did it take Shelbie to run that distance?

1 answer:

castortr0y [4]3 years ago

4 0

Answer:

12.5 seconds

Explanation:

If she runs 1.2 meters per second, and she goes 15 meters, then you can solve the problem by solving 15 / 1.2, which is equal to 12.5.

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What is the maximum force that could be applied to anterior cruciate ligament (ACL) if it has a diameter of 4.8 mm and a tensile

vovangra [49]

Answer:

Maximum force, F = 1809.55 N

Explanation:

Given that,

Diameter of the anterior cruciate ligament, d = 4.8 mm

Radius, r = 2.4 mm

The tensile strength of the anterior cruciate ligament, $P=100\times 10^6\ N/m^2=10^8\ Pa$

We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

$F=P\times A\\\\F=10^8\times \pi (2.4\times 10^{-3})^2\\\\F=1809.55\ N$

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N

4 0

4 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

A: How far did she travel? <br><br><br><br> B: How long did she take?

allochka39001 [22]

A. 60 miles
B. 5 hours

Unless you are looking for slope, in which case the answer is different

7 0

3 years ago

Which of the following is not an example of a physical change?

Mazyrski [523]

Answer:

Explanation:

Cutting a string in half because

b is irreversible

c is a cheical and d is also a chemical change

8 0

3 years ago

A scientist plans to find out the percent of teenagers who like science. She interviews 500 teenagers leaving a science museum a

Georgia [21]

The source of information was biased. It was like walking along a river bank in the country and asking everybody you meet whether they like fishing. Or asking 500 people sitting in the bleachers whether they like baseball.
I'm sure the scientist would have gotten different data if she interviewed 500 teenagers at neighborhood basketball courts, or 500 teenagers at a rock concert.

5 0

3 years ago

Read 2 more answers