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2 years ago

The weight of an objects can be calculated by multiplying mass by

Physics

2 answers:

Lana71 [14]2 years ago

7 0

Acceleration due to gravity.

dmitriy555 [2]2 years ago

6 0

It can be calculated by multiplying mass and free-fall acceleration

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The line graph below shows the number of downloads of two songs after

daser333 [38]

Answer: D

1200

Explanation:

Song 1 is spotted with a cube sign.

At 3 minute, trace the spot to the vertical axis. And you will notice that it a little bit above 10.

Since it is above 10, let assume it is equal to 12.

The number of song downloaded are in hundreds. Therefore, multiply the 12 by 100

12 × 100 = 1200 downloads

Approximately, song 1 has 1200 downloads at minute 3

5 0

2 years ago

Does the image above correctly illustrate how the coriolis effect impacts global winds?

Sedbober [7]

Yes the winds are moving in a straight line

6 0

2 years ago

ELECTROMAGNETIC FIELDS PLAN AN INVESTIGATION LAB REPORT FOR EDG.!!!!! WORTH 100 POINTS????????

tekilochka [14]

Answer:ummm ok

Explanation:I really don’t get it but ok

3 0

2 years ago

A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the

svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength is given by

$\lambda = \dfrac{2L}{n}$

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

$f = n\dfrac{v}{2L}$

now, wavelength calculation

n = 1

$\lambda_1 = \dfrac{2\times 2}{1}$

λ₁ = 4 m

n = 2

$\lambda_2 = \dfrac{2\times 2}{2}$

λ₂ = 2 m

n =3

$\lambda_3 = \dfrac{2\times 2}{3}$

λ₃ = 1.333 m

now, frequency calculation

n = 1

$f = n\dfrac{v}{2L}$

$f_1 =1\times \dfrac{50}{2\times 2}$

f₁ = 12.5 Hz

n = 2

$f = n\dfrac{v}{2L}$

$f_2 =2\times \dfrac{50}{2\times 2}$

f₂= 25 Hz

n = 3

$f = n\dfrac{v}{2L}$

$f_3 =3\times \dfrac{50}{2\times 2}$

f₃ = 37.5 Hz

8 0

3 years ago

In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on

vekshin1

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

$dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J$

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

3 0

2 years ago