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maks197457 [2]
2 years ago
8

A stool has a round non-slip pad fitted to the bottom of each leg.

Physics
1 answer:
Andreyy892 years ago
3 0
50 N/cm^2. It is 75+525/4x3. Pls mark me brainliest
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Where are the longest continuous mountain ranges on Earth located?
andrew-mc [135]
It is located through seven countries in South America which are Venezuela, Colombia, Ecuador, Peru, Bolivia, Chile and Argentina. This is the Andes Mountains which is about 7000 km long. This is known as the world's longest mountain range.
6 0
3 years ago
Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max
yanalaym [24]

Answer:

\theta=39.49^{\circ}

Explanation:

Maximum height of the pumpkin, H_{max}=9.99\ m

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

H_{max}=\dfrac{v^2\ sin^2\theta}{2g}

On rearranging the above equation, to find the angle as :

\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})

\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})

\theta=39.49^{\circ}

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0
2 years ago
If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply
Flauer [41]

Answer:

T=8.33A

Explanation:

From the question we are told that:

Number of battery n=10

Voltage sourceE=12V

Lamp PowerP=10W

Generally the equation for Resistance is mathematically given by

 R=\frac{V^2}{P}

 R=\frac{12^2}{10}

 R=14.4ohms

Therefore

 R_{eq}=\frac{14.4}{10}

 R_{eq}=1.44

Generally the equation for Current is mathematically given by

 T=\ffrac{V}{Req}

 T=\frac{12}{1.44}

 T=8.33A

6 0
2 years ago
A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o
Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let q_1 and q_2 be the charges such that

q_1+q_2=4.7

and Force between charge particles is given by

F=\frac{kq_1q_2}{r^2}

4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}

q_1\cdot q_2=0.522

put the value of q_1

q_2\left ( 4.7-q_2\right )=0.522

q_2^2-4.7q_2+0.522=0

q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}

q_2=0.114 C

thus q_1=4.586 C

3 0
3 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
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