Ask question

Ask question

All categories

2 years ago

10

A projectile is launched straight upward at 50 m/s. Neglecting air resistance its speed upon returning to its starting point is:

________.

1 answer:

lbvjy [14]2 years ago

4 0

Answer:

The speed of the projectile upon returning to its starting point is 50 m/s.

Explanation:

Given;

initial velocity of the projectile, u = 50 m/s

The velocity of the projectile is maximum before hitting the ground. As the object moves upward, the velocity reduces as it approaches the maximum height, at the maximum height the velocity becomes zero. Also, as the object moves downward, the velocity starts to increase and becomes maximum before hitting the ground.

Therefore, the speed of the projectile upon returning to its starting point is 50 m/s.

You might be interested in

Ksivusya [100]

5.77 × $10^1^4$ Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

c = Speed of light = 3 × $10^8$ m/s

λ = Wavelength of light

∴ f = c / λ

f = $\frac{3*10^8}{520 * 10^-^9}$

= 5.77 × $10^1^4$ Hz

Therefore, 5.77 × $10^1^4$ Hz is the green photon's frequency .

Learn more about wavelength here:

brainly.com/question/10728818

#SPJ1

4 0

1 year ago

A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true o

madam [21]

Answer:

D. The motion cannot be determined without knowing the speeds of the objects before the collision.

Explanation:

This question is tricky! We know the object moving to the left has a greater mass than the one moving to the right. We'd <em>assume</em> they would move to the left because the leftwards object has a greater mass, right?

Not. So. Fast.

We can solve for the objects' final velocity using the formula for momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v .

Now here's where the trap is sprung: <em>we don't think about the equation</em>. This shows that the final velocity of the objects and the direction depends on both the mass of the objects <em>and</em> their initial velocity.

Basically, what if the 3 kg object is moving at 1 m/s and the 4 kg object is moving at –0.5 m/s? The objects would move to the <em>right</em> after the collision!

Do we know the velocity of these objects? No, right?

That means we <em>can't</em> determine the direction of their motion <u>unless we know their initial, pre-collision velocity</u>. This question is tricky because we look at the 4 kg vs. 3 kg and automatically assume the 4 kg object would dictate the direction of motion. That's not true. It depends on velocity as well.

I hope this helps you! Have a great day!

4 0

3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

2 years ago

A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-

kifflom [539]

Magnitude of the force of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

The force of tension, $F_T$ , acting up along the plane
The force of friction, $F_f=14.8 N$ , acting down along the plane
The component of the weight in the x-direction, $F_{gx}$ , acting down along the plane

We know that the magnitude of the weight is

$F_g=70.0 N$

So its x-component is

$F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N$

The net force along the x-direction can be written as

$F_x = F_T-F_f-F_{gx}$

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

$F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N$

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0

3 years ago

Read 2 more answers

A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec

horsena [70]

WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J

5 0

2 years ago