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natali 33 [55]
3 years ago
6

A 0.0010-kg pellet is fired at a speed of 50.0m/s at a motionless 0.35-kg piece of balsa wood. When the 

Physics
2 answers:
aleksandr82 [10.1K]3 years ago
8 0
P = m*v

conservation of momentum suggests

initial momentum equals final momentum

mv-initial = mv-final

(0.0010 kg)(50 m/s) = (0.0010 kg + 0.35 kg)v

thus:

v = (0.0010)(50)/(0.351) = 0.142 m/s
vladimir1956 [14]3 years ago
8 0

Answer: The pellet and wood slide at a speed of 0.142 m/s.

Explanation:

To calculate the velocity of the pellet and wood after the collision, we use the equation of law of conservation of momentum, which is:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1 = mass of pellet = 0.001kg

u_1 = Initial velocity of pellet = 50m/s

v_1 = Final velocity of pellet = v m/s

m_2 = mass of wood = 0.36kg

u_2 = Initial velocity of wood = 0m/s

v_2 = Final velocity of wood = v m/s

Putting values in above equation, we get:

(0.001\times 50)+(0.35\times 0)=(0.001+0.35)v\\\\v=0.142m/s

Hence, the pellet and wood slide at a speed of 0.142 m/s.

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Suppose a parachutist is falling toward the ground, and the downward force of gravity is exactly equal to the upward force of ai
Serhud [2]

Answer:

Option b. is correct.

Explanation:

In the given question a parachutist is falling toward the ground .

Also, the downward force of gravity is exactly equal to the upward force of air resistance.

So, net force applied to the parachutist is equal to zero ( because both force acts in opposite direction ).

Now by first law of motion  :

An object will be in rest or in constant speed unless and until no external force is applied on it .

So, in the question the velocity of the parachutist is not changing with time.

Therefore, option b. is correct.

Hence, this is the required solution.

7 0
3 years ago
A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

x_t(t)=d+\frac{1}{2}a_tt^2

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

a_t=2.10 m/s^2 is the acceleration of the truck

The car position instead it is given by the equation

x_c(t)=\frac{1}{2}a_ct^2

where

a_c=3.40 m/s^2 is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

x_t(t') = d + 60

Therefore, solving the equation, we find the time t when  this occurs:

d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

x_c(t')=x_t(t')

And by solving the equation, we find the value of d, the initial distance between car and truck:

\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m

C)

In order to find the speed of each vehicle, we use the following suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

a_t = 2.10 m/s^2

So its speed after t = 7.56 s is

v_t = 0+(2.10)(7.56)=15.9 m/s

For the car, we have

u = 0

a_c=3.40 m/s^2

So its speed after t = 7.56 s is

v_c=0+(3.40)(7.56)=25.7 m/s

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
A 34.0 %-efficient electric power plant produces 800 MW of electric power and discharges waste heat into 20∘C ocean water. Suppo
mojhsa [17]

Answer:

77647

Explanation:

\eta = Efficiency = 34%

Power used in 1 home = 0.02 MW

Total power is

P=\dfrac{800}{\eta}\\\Rightarrow P=\dfrac{800}{0.34}\\\Rightarrow P=2352.94117\ MW

Waste of power

2352.94117-800=1552.94117\ W

Number of homes would be given by

n=\dfrac{1552.94117}{0.02}=77647.0585\ homes

The number of homes that could be heated with the waste heat of this one power plant is 77647

8 0
3 years ago
A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0
3 years ago
Please help me with the question below
vampirchik [111]

Answer:

a cold air mass and a warm air mass merge together

6 0
2 years ago
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