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2 years ago

8

Zoe is shown two mystery boxes that are both 8,000cm3. Her teacher tells her that one mystery box is filled with rocks and the o

ther is filled with napkins. Without touching the boxes, Zoe is asked to determine which box has the napkins.

1 answer:

krek1111 [17]2 years ago

5 0

Answer:

The box of rocks will have depression which can be seen without touching the box.

Explanation:

The density of rocks is very large as compared with napkins. So, the weight of the rocks will be much more greater than that of napkins.

As both boxes have same volume the heavier box will show depression on the lower surface as compared to the lighter box. So, the box of rocks will have depression which can be seen without touching the box.

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The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit

Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V= 30 m/s ( + x direction )

d) λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

As we know that standard form of wave equation given as

$y=A sin(\phi -\omega t)$

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s ( + x direction )

V = f λ

30 = 95.49 x λ

λ = 0.31 m

We know that speed is the rate of displacement

$U=\dfrac{dy}{dt}$

$U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

$U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s$

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0

3 years ago

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$

$\frac {12.8 \ kg*m/s^2}{2 \ kg}=a$

The units of kilograms cancel.

$\frac {12.8}{2}\ m/s^2=a$

$6.4 \ m/s^2=a$

The acceleration is 6.4 meters per square second.

4 0

2 years ago

(15 points) (Asap!!)<br>In what two ways can you increase the elastic potential energy of a spring?

adelina 88 [10]

Hello There

Answers: T<span>he elastic potential energy can be increased by: </span>

<span>1) Getting a spring with a higher spring constant</span>

<span>2) Increasing the length at which the spring is compressed.

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris</span>

5 0

2 years ago

Read 2 more answers

which do you think would be warmer on a winter day when there is no wind a thick forest or a grassy field? explain your answer

Tju [1.3M]

I think it would be warmer in a grassy field with no wind on a winter day because you'll have sunlight hitting you. However, if you were in a thick forest, all sunlight would be blocked and you would have no warmth from the sun.

4 0

3 years ago

Read 2 more answers

Two airplanes leave an airport at the same time.The velocity of the first airplane is 700 m/h at a heading of 31.3 the velocity

MArishka [77]

Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

$v_{12} = v_1 - v_2$

here

speed of first plane is 700 mi/h at 31.3 degree

$v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j$

$v_1 = 598.12\hat i + 363.7\hat j$

speed of second plane is 570 mi/h at 134 degree

$v_2 = 570 cos134 \hat i + 570 sin134 \hat j$

$v_2 = -396\hat i + 410\hat j$

now the relative velocity is given as

$v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j$

$v_{12} =994.12\hat i -46.3 \hat j$

now the distance between them is given as

$d = v* t$

$d = (994.12 \hat i - 46.3 \hat j)* 3$

$d = 2982.36\hat i - 138.9\hat j$

so the magnitude of the distance is given as

$d = \sqrt{2982.36^2 + 138.9^2}$

$d = 2985.6$ miles

so the distance between them is 2985.6 miles

5 0

2 years ago

Read 2 more answers