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Dima020 [189]
3 years ago
8

A student throws a pumpkin up from 14.2 meters with a speed of 19.4m/s. what is the velocity

Physics
1 answer:
RSB [31]3 years ago
6 0

.73 because velocity is determined by distance divide by speed brainliest if this helps


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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

5 0
3 years ago
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t
viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

mv+MV=(m+M)v'

Here , v' is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s

Hence , this is the required solution .

3 0
3 years ago
Which graph shows an object that is dropped?
Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0
3 years ago
Convert 500ml to how many liters
Mashutka [201]

500 ml = 0.5 liters. that's what i'm getting

hope it helps

7 0
3 years ago
Read 2 more answers
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