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3 years ago

Why is it important to have a control setup in an experiment?

Physics

1 answer:

Lana71 [14]3 years ago

5 0

Explanation:

A control is important for an experiment because it allows the experiment to minimize the changes in all other variables except the one being tested.

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What is the density of 15kg block of wood wuth a volume of 1.5 cubic meters<br>

Temka [501]

Answer:

The formula for density is d=m/v so I divided 15 by 1.5 and it gave me 10

Explanation:

8 0

3 years ago

Read 2 more answers

Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the

Hoochie [10]

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

Final position: Mr. Llama's house;
Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

7 0

4 years ago

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration

miss Akunina [59]

Answer:

1728.42857143 m/s²

0.00155883061577 s

259.264285715 m/s²

0.0103922041051 s

The child will get injured if he/she falls on a hardwood floor

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2gs\\\Rightarrow v=\sqrt{2gs+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.37+0^2}\\\Rightarrow v=2.69432737432\ m/s$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-2.69432737432^2}{2\times 2.1\times 10^{-3}}\\\Rightarrow a=-1728.42857143\ m/s^2$

Magnitude of deceleration is 1728.42857143 m/s²

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-2.69432737432}{-1728.42857143}\\\Rightarrow t=0.00155883061577\ s$

Time taken is 0.00155883061577 s

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-2.69432737432^2}{2\times 1.4\times 10^{-2}}\\\Rightarrow a=-259.264285715\ m/s^2$

Magnitude of deceleration is 259.264285715 m/s²

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-2.69432737432}{-259.264285715}\\\Rightarrow t=0.0103922041051\ s$

Time taken is 0.0103922041051 s

It is likely that the child will get injured if he/she falls on a hardwood floor.

It is less likely that the child will get injured if he/she falls on a carpeted floor.

7 0

4 years ago

Circle the anomalous result in the table below. An anomaly is a piece of data that doesn't

alexandr1967 [171]

Answer:

Explanation:

yes

3 0

1 year ago

A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as a box car passes him, the fugitive

Dima020 [189]

Answer:Given:

Initial speed of fugitive, v0 = 0 m/s

Final speed, vf = 6.1 m/s

acceleration, a = 1.4 m/s^2

Speed of train, v = 5.0 m/s

Solution:

t = (vf-v0)/a

t = (6.1-0)/1.4

t =4.36 s

Distance traveled by train, x_T =v*t

x_T =5*4.36 = 21.8 m

Distance travelled by fugitive, x_f = v0*t+1/2at^2

x_f = 0*4.36+1/2*1.4*4.36^2

x_f =13.31 m

5*t = v(t-4.36)+x_f

5*t=6.1*(t-4.36)+13.31

solve for t, we get

t = 12.08 s

The fugitive takes 12.08 s to catch up to the empty box car.

Distance traveled to reach the box car is

X_T = v*t

X_T = 5*12.08 s

X_T = 60.4 m

Explanation:

5 0

3 years ago