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A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860

Charra [1.4K]

Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

Momentum=0.162 x 860

Momentum=139.32kgm/s

3 0

2 years ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

2 years ago

Read 2 more answers

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

5 0

2 years ago

A 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what i

never [62]

Answer:

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p) -

$p_{i}$ = $p_{f}$

System initially is at rest

∴ $p_{i}$ =0

∴ From the above 2 equations

$p_{f}$ =0

We know that ,

Momentum(p)=Mass of the body(m)×velocity of the body(v)

Let $m_{1}$ and $m_{2}$ be the mass of the person and the book respectively and $v_{1}$ and $v_{2}$ be the final velocities of the person and book respectively.

∴ $p_{f}$ = $m_{1}$ $v_{1}$ + $m_{2}$ $v_{2}$ =0

From the question ,

$m_{1}$ = 74.9 kg

$m_{2}$ = 2.44 kg

$v_{2}$ = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × $v_{1}$ )+ (2.44×8.25) = 0

∴ $v_{1}$ = - 0.2687 m/s (Negative sign suggests that the motion of the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

4 0

2 years ago

pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

solution:

given:

p_o=1.07*10^5 N/m^2

ρ_infinity=1.23 kg/m^2

p_infinity=1.01*10^5 N/m^2

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0

2 years ago