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m_a_m_a [10]
3 years ago
11

If you travel for three hours of a speed of 30 km/h, how far will you go?

Physics
1 answer:
lesya692 [45]3 years ago
4 0

Answer:

90km

Explanation:

every hour you travel 30km and you are traveling 3 hours 30×3

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The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventuall
ozzi

Answer:

For a (1) net force will be greater than the weight of the ball

For b (2) net force will be lesser than the weight of the ball

Explanation:

For (a):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes upward, one thing is for sure, the net force is greater than the weight of the ball, because three forces are applied during upward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is also acting downward as it is creating friction between ball and air molecules, so creating hindrance in upward motion

External force to throw ball upward

So

Net Force = Upward force - Air friction - Weight

Since ball is going upward, so net force is greater than both weight and air friction which are pulling ball downward.

For (b):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes downward, one thing is for sure, the net force is lesser than the weight of the ball, because two forces are applied during downward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is acting upward as it is creating friction between ball and air molecules, so creating hindrance in downward motion

So

Net Force = Weight - Air friction

Since ball is going downward, so weight is greater than net force which is in this case is air friction which is pulling ball upward.

4 0
3 years ago
A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and
Angelina_Jolie [31]

A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m

net work done by gravity on the cyclist = mass * gravity * height diff.

= 85 * 9.8 * 12.56

= 10470J

= 10.5kJ

B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J

vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

5 0
3 years ago
What did Bohr’s model of the atom include that Rutherford’s model did not have?
hammer [34]
Bohr explained that the electron revolve around a orbit of fixed energy level that Rutherford fails to explain !!!
8 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
xxMikexx [17]
14-6 =8
8/4= 2m/s per second
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3 years ago
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An external resistor with resistance R is connected to a battery that has an emf E and an internal resistance r. Let P be the el
satela [25.4K]

Answer:

a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes

Explanation:

<u>Solution  :</u>

(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation  in the next form  

P=∈*I-I^2*r                (1)

Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by  

I=∈/R+r                     (2)

When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

I= ∈/r

Now let us plug this expression of I into equation (1) to get the consumed power  

P=∈*I-I^2*r

 =I(∈-I*r)

 =0

The consumed power when R is very small is zero  

(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes  

I=∈/R

The dissipated power due toll could be calculated by using equation.

P=I^2*r                (3)

Now let us plug the expression of I into equation (3) to get P  

P=I^2*R=(∈/R)^2*R

 =∈^2/R

4 0
3 years ago
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