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3 years ago

If you fire a gun horizontally, and at the same time drop a bullet/slug from the same height, what will happen?

Physics

1 answer:

Elena-2011 [213]3 years ago

8 0

Answer:

There you have it. The y-component of air resistance for the fired bullet still depends on the fired speed of the bullet (since it is proportional to v2). A fired bullet (with air resistance) does not hit the ground at the same time as a dropped bullet.

Explanation:

hope this helped ✨

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A 15kg dog jumps out of a 40kg canoe. If the dogs velocity is 1.2m/s, what is the velocity of the canoe?

vivado [14]

Answer:

v = -0.45 m/s

Explanation:

Assuming the canoe was initially at rest with momentum L = 0

and that the dog's velocity is in the positive direction

conservation of momentum

0 = 15(1.2) + 40v

v = -0.45 m/s

6 0

2 years ago

Explain how you can work out the

Ne4ueva [31]

Answer:

The speed of light changes as it moves between media. This causes refraction. Angles of refraction can be calculated using known speeds or wavelengths. Beyond the critical angle, light is reflected.

8 0

2 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0

2 years ago

Can someone help me with one through seven I will mark you the brainly

ki77a [65]

Answer:

1. F = M x A

2. Force

3. 2nd Law: Force

4. a, b, c (in order)

5. 3rd Law: Action and Reaction

6. b, c, a (in order)

7. 1st Law: Inertia

3 0

2 years ago

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When is the magnitude of the acceleration of a mass on a spring at its maximum value?

Andreas93 [3]

Answer:

A. when the mass has a speed of zero

Explanation:

In mass-spring system, the velocity and the acceleration are in anti-phase, which means that when one of the two quantities is maximum, the other one is zero, and vice-versa.

In fact:

- When the displacement of the spring is zero (x=0), the velocity is maximum, due to conservation of energy. In fact, as the displacement is zero, the elastic potential energy of the system (given by $\frac{1}{2}kx^2$ ) is zero, therefore the kinetic energy (given by $\frac{1}{2}mv^2$ ) must be maximum, and so the velocity (v) is also maximum. On the cotnrary, acceleration (a) is directly proportional to the restoring force of the spring, given by

$F=-kx$

so we see that when x=0, then the force is zero: F=0, and so the acceleration is zero as well.

- When the displacement of the spring is maximum, the velocity is zero, due to conservation of energy. In fact, as the displacement is maximum, the elastic potential energy of the system (given by $\frac{1}{2}kx^2$ ) is maximum, therefore the kinetic energy (given by $\frac{1}{2}mv^2$ ) must be zero, and so the velocity (v) is also zero. On the cotnrary, since acceleration (a) is directly proportional to the restoring force of the spring, given by

$F=-kx$

so we see that when x=maximum, then the force is maximum, and so the acceleration is maximum as well.

Based on this, the correct answer is

A. when the mass has a speed of zero

5 0

2 years ago

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