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kap26 [50]
3 years ago
10

A projectile is launched straight upward at 50 m/s. Neglecting air resistance its speed upon returning to its starting point is:

________.
Physics
1 answer:
lbvjy [14]3 years ago
4 0

Answer:

The speed of the projectile upon returning to its starting point is 50 m/s.

Explanation:

Given;

initial velocity of the projectile, u = 50 m/s

The velocity of the projectile is maximum before hitting the ground. As the object moves upward, the velocity reduces as it approaches the maximum height, at the maximum height the velocity becomes zero. Also, as the object moves downward, the velocity starts to increase and becomes maximum before hitting the ground.

Therefore, the speed of the projectile upon returning to its starting point is 50 m/s.

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4. A force of 5 N gives a mass m,, an acceleration of 10 m's, and a mass
Art [367]

Answer:

For mass m  1  newton 2nd law

F=m  1  a  1

​5=m  1  ×10

m  1  =  2 1 kg

For mass m  2

​F=m  2  a 2

​5=m  2 ×20

m  2 =  4 1  kg

if tied together  

Total mass =m  1  +m  2  =  1/2 +1/4=3/4kg  

Now

F=M T  Q T

 a  T    =  5/m T =  5×4/3  =  3 /20m/s^2

Explanation:

8 0
3 years ago
What is the volume of this bubble when it reaches the surface?
steposvetlana [31]

Answer:

Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.

8 0
3 years ago
A 3.0 kg ball is lifted two meters upward in two seconds, and a 6.0 kg ball is lifted one meter upward in one second? Which requ
Sav [38]

Answer:

1. They both uses same energy

2. The 6 kg ball requires more power than 3kg ball

Explanation:

Sample 1

m = 3kg

g= 10m/s^2

h = 2m

t = 2secs

W = mgh = 3 x 10 x 2 = 60J

P= w/t = 60/2 = 30watts

Sample 2

m = 6kg

g= 10m/s^2

h = 1m

t = 1sec

W = mgh = 6 x 10 x 1 = 60J

P= w/t = 60/1 = 60watts

They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball

3 0
4 years ago
Two charges 5 c and 15 c are seperated by 10 cm. what is the electric force between them?
icang [17]

Answer:

6.75×10^13N

Explanation:

The electric force between the charges can be determined using coulombs law which states that 'the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of their distance between them.

Mathematically, F = kq1q2/r² where;

q1 and q2 are the charges

r is the distance between the charges

F is the force of attraction

k is the coulombs constant

Given q1 = 5C q2 = 15C r = 10cm = 0.1m k = 9×10^9Nm²/C²

Substituting the given values in the formula we have;

F =9×10^9×5×15/0.1²

F = 6.75×10^11/0.01

F = 6.75×10^13N

Therefore the electric force between them is 6.75×10^13N

8 0
3 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

4 0
4 years ago
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