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2 years ago

Please help!

Physics

1 answer:

julsineya [31]2 years ago

6 0

Answer:

Answer:

3 g/cm

Explanation:

Mass of stone = 30 g

Initial volume of cylinder = 50 cm

Increased volume = 60 cm

Volume = 60-50

= 10 cm

Density of stone = Mass/Volume

= 30/10

= 3 g/ml as 1 ml = 1 cm

= 3 g/ml

Explanation:

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An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0

3 years ago

Which of the following statements about ycarrier(x,t) is correct?

finlep [7]

Answer: Option D : is traveling rapidly but oscillating slowly.

Explanation:

ycarrier(x,t) is traveling rapidly but oscillating slowly.

8 0

3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

Why is friction like applied force but different from gravity?

kirill115 [55]

Answer:

Friction is when a force is applied or done by weight dragging onto something.

Explanation:

Gravity is when an object is getting pulled toward the center of what is attracting it. And applied force is when someone/sommething is applying force.

3 0

3 years ago

Read 2 more answers

Pleaseeee Please help, I will love you forever and ever

matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J = - 63000 J

Cp = 4184 J / kg°C

Formula

Q = mCp(T2 - T1)

T2 = T1 + Q/mCp

Substitution

T2 = 35 - 63000/(0.5 x 4184)

T2 = 35 - 63000/2092

T2 = 35 - 30.1

T2 = 4.9 °C

6 0

3 years ago