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klio [65]
3 years ago
5

Plz explain the answers with steps and a little bit of explanation. I really need the answers for my exam tomorrow! Please help

me!

Physics
1 answer:
DochEvi [55]3 years ago
7 0
Here's what you need to memorize for your exam tomorrow.

Distance = (speed) x (time) .

That's it. Memorize it.

-- If the question wants you to find speed, use it exactly in that form.

-- If the question wants you to find speed, then divide each side by (time)
and it says

. Time = (distance) / (speed) .

-- If the question wants you to find time, then divide each side by speed,
and it says

. Time = (distance) / (speed) .

So if you memorize that one equation ... Distance = (speed) x (time) ...
you can solve ANY problem to find distance, speed, or time.

On the sheet in the picture . . . . .

#2). The question is "How long ?". That's TIME that you have to find.
Use the equation in the form of

. TIME = (distance) / (speed)

. = (60 km) / (48 km/h)

. = 1.25 hours .

#3). This one wants you to find SPEED. Use the equation in the form of

. SPEED = (distance) / (time)

but be careful. The time has to be in hours. 55 minutes = 55/60 of an hour.

. SPEED = (distance) / (time)

. = (60 km) / (55/60 hour)

. = (60 x 60 km) / (55 hour)

. = 65.45 km/hr .

#4). This one wants you to find TIME. (It says "How long ?".)

It's two trips, so you have to find the time for each trip.

First trip: TIME = (distance)/(speed) = (24 km)/(65 km/hr) = 0.369 hr

Second trip: TIME = (distance)/(speed) = (50 km)/(80 km/hr) = 0.625 hr

Total time for both trips = (0.369 hr) + (0.625 hr) = 0.994 hour
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fiasKO [112]

Answer:

B. 1500 kg*m/s

Explanation:

Momentum p = m* v

In any type of collision, the total momentum is preserved!

The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.

p1 + p2 =

m1*v1 + m2*v2

m1 = me = 300 kg

v1 = 3 m/s

v2 = 2 m/s

Substitute the given numbers:

300*3 + 300+2

900 + 600

1500 kg*m/s, which is answer B.

3 0
3 years ago
A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially
Alchen [17]

The new speed of car is 10.9 m/s

<h3 />

According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

m_1u_1 + m_2u_2 = (m_1 + m_2) v

v =\frac{m_1u_1}{m_1+m_2}

v = \frac{7950*15}{7950+2950}

v= 10.9 m/s

Therefore, the new speed of care is 10.9 m/s

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2 years ago
In order for a color to be seen by a human, the light must be
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In the part of the spectrum our eyes can detect (a spectrum is an arry of entities, as light waves or particles, ordered in accordance with the magnitudes of a common physical property, as wavelength or mass) Hope this helps you :D
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Light is an electromagnetic wave and travels at the speed of 3.00 x 10^8 m/s
Over [174]
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6 0
3 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
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