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3 years ago

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Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the b

all?

1 answer:

Ghella [55]3 years ago

3 0

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:

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A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

3 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

6 0

3 years ago

Non examples of gravity ?

Afina-wow [57]

Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.

3 0

3 years ago

Read 2 more answers

Planets move around the sun in elliptical orbits. True. False

astra-53 [7]

this answer is true.

5 0

3 years ago

Read 2 more answers

Which describes the phenotype of the tomato plant

blagie [28]

The allele for red skin is dominant R
The allele for yellow skin is recessive r
The genotype Rr is heterozygous, resulting in phenotype: red skin tomato
The genotype RR is dominant homozygous, resulting in phenotype: red skin tomato
The genotype rr is recessive homozygous, resulting in phenotype: yellow skin tomato

Read more on Brainly.com - brainly.com/question/10183300#readmore

4 0

3 years ago