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Illusion [34]
2 years ago
12

How much work is it to lift a 20. kg sack of potatoes vertically 6.5 m?

Physics
1 answer:
Dvinal [7]2 years ago
4 0
W=FD
W=20*6.5
W= 130 Joul
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A 31 kg crate full of very cute kittens is placed on an incline that is 17° below the horizontal. The crate is connected to a sp
fgiga [73]

Answer:

Explanation:

Change in length of spring = 2.13 m

Component of weight acting on spring = mg sinθ

so

mg sinθ = k x where k is spring constant and x is total stretch due to force on the spring.

Here x = 2.13

mg sin17 = k x 2.13

31 x 9.8 sin17 = k x 2.13

k = 41.7 N/m

b ) In case surface had friction , spring would have stretched by less distance .

It is so because , the work done by gravity in stretching down is stored as potential energy in  spring . In case of dissipative force like friction , it also takes up some energy in the form of heat etc  so spring stretches less.

5 0
3 years ago
A branch falls from a tree How fast is the branch moving after 0 28 seconds
egoroff_w [7]

Answer:

c. 2.7 m/s

Explanation:

v = a.t + v_{0}\\a = g = 9.81 m/s^{2}} \\t=0.28 s \\v_{0} = 0\\=> v = 9.81 * 0.28 = 2.74 m/s

3 0
3 years ago
The wave shown below is produced in a rope.
Kay [80]

Answer:

<h3>B) A dot vector drawn up and vector drawn down.</h3>

Explanation:

Brainliest?

6 0
3 years ago
Read 2 more answers
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
salantis [7]

Answer:

Q = 4.63 \times 10^6 J

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

s = 240 J/kg C

now we will have

Q = ms\Delta T + mL

\Delta T = 961.8 - 20

\Delta T = 941.8 degree C

now from above equation

Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)

Q = 3.1 \times 10^6 + 1.53 \times 10^6

Q = 4.63 \times 10^6 J

3 0
3 years ago
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