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gregori [183]
3 years ago
5

An arrow of mass 415 g is shot at a target with a speed of 68.5m/s. The target, which has a mass of 3.3 kg, is moving toward the

arrow with a speed of 1.1m/s. The surface on which the target is sliding is friction-free. If the arrow embeds itself in the target, what will be the velocity of the target/arrow mass after the collision? Take the arrow's initial velocity direction as positive
Physics
1 answer:
Murrr4er [49]3 years ago
4 0

The velocity of the target and arrow after collision is 6.67m/s

<u>Explanation:</u>

Given:

Mass of arrow, mₐ = 415g

Speed of arrow, vₐ = 68.5m/s

Mass of the target, mₓ = 3.3kg = 3300g

speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction

Velocity of the target and arrow after collision, vₙ = ?

Applying the conservation of momentum,

mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ

415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ

28427.5 - 3630 = 3715 X vₙ

24797.5 = 3715 X vₙ

vₙ = 6.67m/s

Therefore, the velocity of the target and arrow after collision is 6.67m/s

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Which of the
kompoz [17]
C. Amount of oxygen

The others either change but don’t decrease or they increase.
8 0
2 years ago
The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.2 × 10-15 m. The single electron in a hydrogen
Nutka1998 [239]

Answer: 0.86 × 10^14

Explanation:

Given the following :

Radius of proton = 1.2 × 10-15 m

Radius of hydrogen atom = 5.3 × 10-11 m

Density of proton could be calculated thus:

Mass of proton = 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45

Density = mass / volume

Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)

= 0.2306 × 10^18

Density of hydrogen atom:

Mass of hydrogen atom= 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31

Density = mass / volume

Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)

= 0.2676 × 10^4

Ratio is thus:

Density of proton / density of hydrogen atom

0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14

6 0
3 years ago
A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant
nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

6 0
2 years ago
Read 2 more answers
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
2 years ago
Can someone help me plzzz..<br>whoever answers the best will be marked as brainliest.....​
IrinaK [193]

Answer:

1) 3 applications of pressure in daily life are :-

● The area of sharp edge of knife, scissor or handsaws are much less then blunt edge. So, for same total force pressure is more for sharp edges than the blunt one. Hence sharp knife, scissors etc, cuts easily than a blunt one.

●Broad handles in bags and suitcases are provided for the comfort. Broad handles have large area. So, the pressure exerted on hands and shoulders would be small while carrying the bags and the suitcases.

●Trucks carrying heavy loads have more than four tyres. More tyres in case of trucks increase the area of contact with the road. This results in reduced pressure on the tyres.

2) Area of the surface which is on ground = 1.5×1

= 1.5m^2

Mass of the block = 300kg

Force applied by the block = Mass × g = 300×10

= 3000N (where g = acceleration due to gravity )

Pressure = Force applied / Area of the surface

= 3000N / 1.5m^2

= 2000 Pa

3)

a) The above experiment signifies that more the area of the surface of an object , less the pressure an object applies.

b) B exerts the minimum pressure because the area of its surface to ground is greater than others & as it has more area of surface , it exerts less pressure. ( area is inversely proportional to pressure )

c) D exerts the maximum pressure because the area of its surface to ground is lesser than others & as it has less area of surface , it exerts more pressure. ( area is inversely proportional to pressure )

d) It depend upon the way an object is kept on ground. If an object is kept in such a way dat the area of the surface to the ground is more , then pressure will be least exerted .If an object is kept in such a way dat the area of the surface to the ground is less, then pressure will be exerted more .

e) Do it yourself . only i will suggest that make the tip of the cone ( which is to the ground ) more narrower.

6 0
3 years ago
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