Ask question

Ask question

All categories

3 years ago

12

10) Allen was curious about dog interaction at the dog park. He wondered if big dogs liked to play with big dogs and if little d

ogs liked to play with little dogs. After gathering information through research, he followed the scientific method by stating a ____________. This educated guess refers back to the original questions and is stated in such a manner that the results can be measured. A) experiment B) hypothesis C) problem D) procedure

1 answer:

GalinKa [24]3 years ago

8 0

Answer: B) hypothesis

Explanation:

You might be interested in

10. A loud motorcycle passes by the front of Bill's house. What medium

garik1379 [7]

Answer: Air

Explanation:

3 0

3 years ago

Read 2 more answers

Melissa's favorite exercise equipment at the gym consists of various springs. in one exercise, she pulls a handle grip attached

konstantin123 [22]

The formula for force exerted on/by a spring is

F = k*e where k is the spring constant and x is the distance stretched from unstrained position. This should allow you to find what you need.

Using F = k x e,

where k is the spring constant,

and e is the extension,

The F is her weight = 45 X 0.80

= 36 N

6 0

3 years ago

What is the momentum of a 45-kg quarterback moving eastward at 15<br> m/s?

puteri [66]

Answer:

Given

mass (m) =45kg

velocity (v) =15m/s

momentum (p) =?

Form

p=mv

=45x 15

p=675kg.m/s

the momentum =675kg.m/s

4 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alys

guapka [62]

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s t=57/0.2 =285 seconds

=4.75 minutes

5 0

3 years ago

Read 2 more answers