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igor_vitrenko [27]
3 years ago
14

Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his

house to talk to a neighbor. It takes Ben 17 minutes from the time he leaves his house until he stops to talk to his neighbor. What is Ben’s average velocity?
Physics
2 answers:
Oksana_A [137]3 years ago
6 0
71 MPM (Meters Per Minute)
S = Speed
D = Distance
T = Time
to find the Speed you divide D by your T
liq [111]3 years ago
6 0

Answer:

0.196 m/s

Explanation:

Average displacement is defined as the total displacement covered to the total time taken.

Total displacement = 500 - 500 - 200 =  - 200 m

Total time taken = 17 minutes = 17 x 60 = 1020 second

So, Average velocity = 200 / 1020 = 0.196 m/s (magnitude only)

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A constant force of 120 N pushes a 55 kg wagon across an 8 m level surface. If the wagon was initially at rest, what is the fina
Illusion [34]

Answer:

The kinetic energy of the wagon is 967.0 J

Explanation:

Given that,

Force = 120 N

Mass = 55 kg

Height = 8 m

We need to calculate the kinetic energy of the wagon

Using newtons law

F = ma

\dfrac{120}{55}=a

a =2.2\ m/s^2

Using equation of motion

v^2 =u^2+2as

Where,

v = final velocity

u = initial velocity

s = height

Put the value in the equation

v^2=0+2\times2.2\times8

v=5.93\ m/s

Now, The kinetic energy is

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times55\times(5.93)^2

K.E=967.0\ J

Hence, The kinetic energy of the wagon is 967.0 J

3 0
3 years ago
State two environmental problems caused by coal fired power stations
Julli [10]

Answer:

Air pollution and water pollution.

Explanation:

The burning of coal causes air pollution, and coal fired power stations dump waste in large bodies of water.

6 0
3 years ago
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.
drek231 [11]

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under  gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I  am  sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

8 0
3 years ago
Fossil fuels like gasoline or coal are examples of
disa [49]

Answer:

Nonrenewable Resources

8 0
3 years ago
A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found
kati45 [8]

Answer:

1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
  • Distance between the rod and the balloon, d = 0.0640 m.
  • Acceleration due to gravity, \rm g = 9.81\ m/s^2.

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

For the elecric force and the weight to be balanced,

\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

3 0
2 years ago
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