4. The value of the before and after-collision momentum of two colliding objects is shown in the

An American Red Cross plane needs to drop emergency supplies to victims of a hurricane in a remote part of the world. The plane

Andreas93 [3]

Refer to the diagram shown below.

Assume that air resistance is ignored.

Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²

The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²

Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s

The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m

Answer: 319.4 m (nearest tenth)

3 0

3 years ago

) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi

natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0

3 years ago

1. If a model train car with a momentum of 12 kg•m/s collides with a 2 kg model train car that is not moving, what is the total

wariber [46]

Answer:

$P_{f} = 12 \ kg.m/s$

Explanation:

Given data:

Momentum of moving model train, $P_{1} = 12 \ kg.m/s$

Mass of the stationary model train, $m = 2 \ kg$

Initial speed of the stationary model train, $v = 0$

Assume there is no external force is acting on the given train system.

In this case, the total linear momentum of the trains would be conserved.

Let the final linear momentum of the trains be $P_{f}.$

Thus,

$P_{i} = P_{f}$

$P_{1} + P_{2} = P_{f}$

$P_{1} + mv = P_{f}$

$12 + 2 \times 0 = P_{f}$

$\Rightarrow \ P_{f} = 12 \ kg.m/s.$

7 0

2 years ago

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

masya89 [10]

Answer: the magnetic wave will travel out of the screen.

Explanation:

Electric field direction is perpendicular to the magnetic field direction. Both are also perpendicular to the direction of the particles.

Using right hand rule to solve this problem,

This pointed finger depicts the electric field direction which the curly fingers depict the direction of the magnetic field. The pointed thumb will depict the direction in which the wave travel. Which is out of the screen.

8 0

3 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

Tcecarenko [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

3 0

3 years ago