All categories

3 years ago

If the mass of a material is 115 grams and the volume of the material is 16 cm3, what would the density of the material be?

1 answer:

hjlf3 years ago

6 0

Well, density is mass/volume. So what's 115 g / 16 cm3? That'll get you your density. Remember that density will be in g/cm3!

You might be interested in

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

german

Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒ $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

6 0

2 years ago

Both a 3 kg book and 5 kg bowling ball are sitting still on the floor. What is true about the kinetic energy of the objects? Exp

Charra [1.4K]

Answer:

The kinetic energy for both objects is the same.

Explanation:

While in other cases the kinetic energies of two objects that have different masses might be different depending on their velocities, in this case both the 3 kg book and 5 kg bowling ball have the same kinetic energy.

This is because kinetic energy is calculated using the formula: K = 1/2 * m * v^2, where m represents the mass and v represents the velocity of the object.

Since the book and the bowling ball are sitting still on the floor, their velocities are zero. Hence, when we plug in 0 for velocity into the equation for kinetic energy, we will get that the kinetic energy is 0 for the book and the bowling ball.

Hope this helps!

6 0

2 years ago

HELLLLLP PLEASE, thank u :)

MrRa [10]

Answer:

Option B. 3.0×10¯¹¹ F.

Explanation:

The following data were obtained from the question:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:

Capacitance (C) = Charge (Q) / Potential difference (V)

C = Q/V

With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

C = Q/V

C = 3.0×10¯⁹ / 100

C = 3.0×10¯¹¹ F.

Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.

7 0

3 years ago

Why are viruses hard to fight

ki77a [65]

Answer:Compared to other pathogens, such as bacteria, viruses are minuscule. And because they have none of the hallmarks of living things — a metabolism or the ability to reproduce on their own, for example — they are harder to target with drugs.

Explanation:

7 0

2 years ago

Does the voltage of a battery affect the strength of an electromagnet?

swat32

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

7 0

2 years ago