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4 years ago

Choose which statements correctly identify the relationship of mass volume and density by clicking the sentence

1 answer:

7 0

Mass is the amount of matter present in an object, it also determines the strength of the mutual gravitational force of an object to another object. Volume is the amount of space that the object occupies. Meanwhile, density is the amount of mass per volume of an object, with that formula, we can say that density is directly proportional to the mass but indirectly proportional to the volume.

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Calculate the amount of charge necessary to exert a force of 35 N in an electric field of 300 N/C.

densk [106]

Answer:

The amount of charge is "0.117 C" i.e., option A.

Explanation:

The given values are:

Force,

$\vec{F}$ = 35 N

Electric field,

$\vec{E}$ = 300 N/C

Charge,

q = ?

As we know,

⇒ $\vec{F}=q\vec{E}$

On substituting the given values in the above formula, we get

⇒ $35=q\times 300$

⇒ $q=\frac{35}{300}$

⇒ $q=0.117 \ C$

3 0

3 years ago

A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After

Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

$v^2 -u^2 = 2ad$

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

$v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s$

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

$d=\frac{1}{2}at^2$

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s$

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

$K_i = U_f\\\frac{1}{2}mu^2 = mgh$

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

$h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m$

So the maximum altitude reached by the rocket is

$h' = d+h=1500 m +550.8 m=2050.8 m$

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

$h=\frac{1}{2}gt^2$

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s$

So the total time of the trip is

$t'=28.9 s+10.6 s=39.5 s$

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

$\frac{1}{2}mv^2 = mgh$

where

v is the final velocity of the rocket

h = 2050.8 m is the initial altitude of the rocket

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(2050.8 m)}=200.5 m/s$

F. 60 s

We need to calculate the time the rocket takes to fall down to the ground from the point of maximum altitude, and that is given by

$h'=\frac{1}{2}gt^2$

where

h' = 2050.8 m

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h'}{g}}=\sqrt{\frac{2(2050.8 m)}{9.8 m/s^2}}=20.5 s$

So the total time of the trip is

$t''=39.5 s+20.5 s=60 s$

6 0

4 years ago

Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not

d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

$I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},$

where $I_0(\text{at }1000\text{ Hz})$ is the hearing threshold at 1000 Hz.

Therefore we have the following

$0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}$

$I_1$ is the threshold at 1000Hz and $I_2$ is the threshold at 100Hz.

By exponentiating we have

$10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.$

Now dividing these two equations we get

$\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.$

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.

7 0

3 years ago

Explain the relationship among visible light, the electromagnetic spectrum, and sight.

yaroslaw [1]

Explanation:

The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.

Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.

8 0

3 years ago

Read 2 more answers

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

Gnom [1K]

Answer:

(a) Acceleration of positron is 6.03 x 10¹³ m/s²

(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s

Explanation:

Given :

Constant electric field, E = 343 N/C

Mass of positron, m = 9.1 x 10⁻³¹ kg

Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C

(a) Coulomb force on the positron is determine by the relation :

F = q x E ....(1)

But, force is also equals to product of mass and acceleration. So,

F = ma .....(2)

Here a is acceleration.

From equation (1) and (2).

m x a = q x E

$a=\frac{qE}{m}$

Substitute the values of q, E and m in the above equation.

$a=\frac{1.6\times10^{-19}\times 343}{9.1\times10^{-31} }$

a = 6.03 x 10¹³ m/s²

(b) Initially, the positron is at rest, so its initial speed is zero.

The equation of motion for positron is :

v = u + at

Here v is final speed, u is initial speed and t is time.

Since, u is zero, so the equation becomes :

v = at

Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.

v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s

v = 5.24 x 10⁵ m/s

6 0

3 years ago