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ElenaW [278]
2 years ago
13

Pls help

Physics
1 answer:
belka [17]2 years ago
4 0

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these  value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

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g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.​
docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

7 0
3 years ago
What is the acceleration of a 349 kg object that moved with a force of 750 N?
zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

From the question we have

a =  \frac{750}{349}  \\  = 2.14899713...

We have the final answer as

<h3>2.15 m/s²</h3>

Hope this helps you

4 0
3 years ago
How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
Three systems of two charged particles are shown below. All the particles have the same mass,
Len [333]

Answer:D

Explanation:

5 0
3 years ago
What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

5 0
3 years ago
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