Ask question

Ask question

All categories

3 years ago

5

what would happen to an object if there were no inertia? Also would this mean that if there were no inertia that an object may t

ravel high speeds DEPENDING on the object's mass?

1 answer:

adelina 88 [10]3 years ago

4 0

There would be infinite acceleration, and the body would have no mass <span />

You might be interested in

1.the student measures the mass of a piece of metal.its mass is 146g.

marissa [1.9K]

Answer:

7.3 g/cubic cm

Explanation:

You divide the mass by the volume to calculate the density of the metal.

146g ÷ 20cm³ = 7.3g/cm³

3 0

2 years ago

Se tienen 500g de alcohol etílico a una temperatura de -40 °C ¿Cuánto calor se necesita para transformarlo a vapor a una tempera

Feliz [49]

The question is: You have 500g of ethyl alcohol at a temperature of -40 ° C. How much heat is needed to transform it into steam at a temperature of 150ºC?

Answer: 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of $-40^{o}C$ to $150^{o}C$ .

Explanation:

Given: Mass = 500 g

Initial temperature = $-40^{o}C$

Final temperature = $150^{o}C$

The standard value of specific heat of ethyl alcohol is $2.46 J/g ^{o}C$ .

Formula used to calculate the heat energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass of substance

C = specific heat

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = m \times C \times (T_{2} - T_{1})\\= 500 g \times 2.46 J/g^{o}C \times [150 - (-40)]^{o}C\\= 233700 J$

Thus, we can conclude that 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of $-40^{o}C$ to $150^{o}C$ .

8 0

2 years ago

A capacitor has a charge of 4.6 Î¼C. An E-field of 1.8 kV/mm is desired between the plates. There's no dielectric. What must be

Scrat [10]

Answer:

$A = 0.2875 m^2$

Explanation:

As we know that

$Q = 4.6 \mu C$

E = 1.8 kV/mm

now we know that electric field between the plated of capacitor is given as

$E = \frac{\sigma}{\epsilon_0}$

now we will have

$1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}$

$\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})$

$\sigma = 1.6 \times 10^{-5} C/m^2$

now we have

$\sigma = \frac{Q}{A}$

now we have area of the plates of capacitor

$A = \frac{Q}{\sigma}$

$A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}$

$A = 0.2875 m^2$

5 0

2 years ago

How do you think air resistance affects measured values of g? If you used a ping pong ball, for example, how would this affect t

valentinak56 [21]

Answer:

A) Air resistance acts in a direction opposite the the fall of an object reducing it by doing work against the weight of the object due to gravity.

B) using a ping pong ball, the time of fall will be greatly reduced since it has little weight (its mass x acceleration due to gravity) against the air resistance. The net downward force of the weight and the air resistance will be small.

C) No, I wouldn't expect them to fall at the same time. The steel ball will have more weight compared to the ping pong ball and hence it will have a larger net force downwards.

D) If they are both released from a 6 m height, the steel ball will fall to the ground first since it has a larger net force downwards.

3 0

2 years ago

A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length

lora16 [44]

Answer:

T = 4.42 10⁴ N

Explanation:

this is a problem of standing waves, let's start with the open tube, to calculate the wavelength

λ = 4L / n n = 1, 3, 5, ... (2n-1)

How the third resonance is excited

m = 3

L = 192 cm = 1.92 m

λ = 4 1.92 / 3

λ = 2.56 m

As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio

v = λ f

f = v / λ

f = 343 / 2.56

f = 133.98 Hz

Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m

The expression for standing waves on a string is

λ = 2L / n

λ = 2 0.37 / 2

λ = 0.37 m

The speed of the wave is

v = λ f

As we have some resonance processes between the string and the tube the frequency is the same

v = 0.37 133.98

v = 49.57 m / s

Let's use the relationship of the speed of the wave with the properties of the string

v = √ T /μ

T = v² μ

T = 49.57² 18

T = 4.42 10⁴ N

4 0

2 years ago

Read 2 more answers