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Leviafan [203]
4 years ago
13

HHEELLPPP ITS ARGENT!!! Just Answer B and C PLEASEEEE

Physics
1 answer:
zavuch27 [327]4 years ago
8 0

Answer:

B.  F = 7.35 N

C.  Equal and opposite

Explanation:

Given

The mass of the football, m = 0.42 Kg

Time of flight of the ball, t = 1.6 s

The initial velocity of the ball, u = 0

The final velocity of ball, v = 28 m/s

The acceleration of the ball is given by the formula

                                       a = (v - u) /t m/s²

Substituting the values in the above equation

                                        a = (28 - 0) / 1.6

                                           = 17.5 m/s²

The force on the ball is given by the formula

                                       F = m x a  newton

Substituting the values in the above equation for force

                                      F = 0.42 Kg x 17.5 m/s²

                                        = 7.35 N

Therefore the force acting on the ball is, F = 7.35 N

According to newton's third law of motion, if a person exerts a force on a ball, the ball also exerts a force that is equal to applied force on the person's foot. It is in a direction opposite to the force exerted by the kicker's foot.

Hence, the forces are equal and acting in opposite direction.

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The distance between two planets is 1600 km. How much time would the light
Snowcat [4.5K]

Answer:

5.33*10^-3 seconds

Explanation:

c = d/t

c = speed of light constant (3.0*10^5 km/s)

d = distance (1600 km)

t = ?

3.0*10^5 = 1600/t

t = 1600/3.0*10^5

t = 5.33*10^-3 seconds

I hope this helped! :)

6 0
3 years ago
A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t
const2013 [10]

Answer:

\theta=12.19^{\circ}

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

sin\theta =\dfrac{u}{v}

Now by putting the values in the above equation we get

sin\theta =\dfrac{30}{142}

sin\theta=0.21

\theta=12.19^{\circ}

Therefore the angle will be 12.19° .

 

4 0
4 years ago
Technician A says that the intake and exhaust manifolds have to be removed before removing the engine from the vehicle. Technici
Amiraneli [1.4K]

Answer: Technician A is correct

Explanation:

The intake manifold is the compactment that all fuel and air supply to the cylinders. It's connected to the engine so it has to be disconnected while the exhaust manifold receives all the exhaust gases from the cylinders and releases the gas through a single or double exhaust gases outlet.

5 0
3 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
3 years ago
An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu
romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

                                   qnet = (Iin - Iout)*t

                                   qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

                                   V = k*qnet / r

Where,                        k = 8.99*10^9   ..... Coulomb's constant

                                   qnet = V*r / k

                                   t = 1000*0.1 / (8.99*10^9 * 0.000002)

                                   t = 5.56 ms

                                   

7 0
3 years ago
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