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azamat
3 years ago
10

A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, w

hat is the density of copper? (hint: find the volume of a block first)
Physics
1 answer:
sp2606 [1]3 years ago
3 0

Answer:

8.9 g/cm^3

Explanation:

density = mass/volume

volume = length * width * height

volume = (8.4 cm)(5.5 cm)(4.6 cm)

volume = 212.52 cm^3

mass = 1896 g

density = (1896 g)/(212.52 cm^3)

density = 8.9 g/cm^3

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Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while
Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is t=2.7\ s

Malda initial velocity u=95\ cm/s

Let h be the height of Cliff

using h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

h=0+\frac{1}{2}(9.8)(2.7)^2

h=35.72\ m

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

R=u\times t

R=0.95\times 2.7=2.43\ m    

7 0
3 years ago
what is the mass of vertical column of air that exerts pressure of 1 atm over a10m by 10m square surface
Oksana_A [137]

Answer:

m=1.01\times 10^6\ kg

Explanation:

Given that,

Pressure, P = 1 atm = 101325 Pa

Area of the square surface, A = 10² = 100 m²

We need to find the mass of vertical column of air. We know that, pressure is equal to the force acting per unit area. So,

P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg

So, the required mass of the vertical column of air is 1.01\times 10^6\ kg.

4 0
3 years ago
4. How long would it take for a water balloon to fall 39 m if we dropped it starting from rest down an
Paraphin [41]

Answer:

Around 2.8212 sec

Explanation:

Given the eqn x=1/2at^2+vot

your vo=0

39=1/2(-9.8)t^2

=7.95=t^2

=2.82sec

8 0
3 years ago
Par 1/2
BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0
3 years ago
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
Anastaziya [24]

Answer:

As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.

This will mean that the denser objects will always go to the bottom.

This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.

There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.

The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.

6 0
3 years ago
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