A 1500 kg car has the same momentum as a 2500 kg truck moving at 25 m/s. How fast is the car moving in m/s

[tex]\huge \bf༆ Answer ༄[/tex]

The Correct choice is ~ A

If an object moves at a constant velocity, then the change in velocity over the time is 0, Acceleration is defined as rate of change in velocity but since there is no change in velocity, the value of Acceleration is equal to Zero.

Explanation:

m = 70 kg

s = 3m

t =2s

g = 10 m/s²

(a)How much work is done by the man?

W = Fs

= mg X s

= 70 x 3 x 10

= 210 x 10

= 2100 Joule

(b)What is the power exerted by the man?

P = W/t

= 2100/2

P = 1050 Watt

Answer:

2.5m/s

Explanation:

m1=55kg

m2=25kg

Ben's velocity before collision=0m/s

Junior's velocity before collision= 8m/s

P1=m1v1+m2v2

P1=55kg(0m/s)+25kg(8m/s)

P1=200kg(m/s)

Know that the momentum before the collision = momentum after the collision

P2=(55kg+25kg)V

200kg(m/s)=(75kg)V

200kg(m/s)/75kg=V

V=2.5m/s.

It makes sense bc after they collide, their speed would slow down, and it wouldn't make sense if the momentum would be greater or less after they collide.

The change in the length of the spring as it stretches from 4.0 cm to 6.0 cm is 2.0 cm.

The change in length of the spring can be calculated by subtracting the initial unstretched length (4.0 cm) from the final stretched length (6.0 cm).

Change in length = Final length - Initial length

Change in length = 6.0 cm - 4.0 cm

Change in length = 2.0 cm

Therefore, the change in the length of the spring as it stretches from 4.0 cm to 6.0 cm is 2.0 cm. This means the spring elongates by 2.0 cm under the applied load between points A and B.

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The complete question is:

The average value of the load between A and B is 6.0 N. The spring has an unstretched length increases from 4.0 cm to 6.0 cm. What is the change in the length of the spring as it stretches from 4.0 cm to 6.0 cm?

The magnitude of the gravitational force the planet exerts on Satellite 2 is  1/2F.

We recall that from Newton's law of universal gravitation, the gravitational force is given by;

F =G m1m2/r^2

m1 and m2 are the respective masses

r is the distance of separation.

Let the mass of the planet be M and the mass of the satellite be m. The distance of separation is r.

For satellite 1;

F = GmM/(2r)^2

F = GmM/4r^2

For satellite 2;

F2= G2mM/r^2

Hence, the magnitude of the gravitational force the planet exerts on Satellite 2 is  1/2F.

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Answer:

Explanation:

Is it a 30 kg ball or a 3 kg ball?

Does not matter as we don't have to do actual calculations

If all of the momentum is transferred out of a mass, the velocity remaining must be ZERO

Answer: they have one path to flow

Explanation: share the same current

Answer:

denser than water:

1. clay

2. coins

3. rock

aren't denser than water

1. wood

2. sponge

3. apple

Answer:

its keeps the fur straight.....im kinda confused with the question

Explanation:

Four regions of the electromagnetic spectrum that astronomers use when observing objects in the space are the following enumerated answers.

1. First is Ultraviolet

2. Next is Infrared

3. Then the radio

4. Lastly the Visible lights.

These are the answers to the problem.

Yes, it's possible to predict the speeds that a coaster will reach before it’s

ever placed on the track.

This is usually calculated with the potential energy which is

Potential energy = m g h

where m is mass, g is acceleration due to gravity and h is height.

The given formula above is used in predicting the speeds that a coaster will

reach before it’s ever placed on the track.

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Answer:

18375g

Explanation:

[tex]\boxed{density = \frac{mass}{volume} }[/tex]

[tex]∴ mass = density \times volume[/tex]

Let's find the volume of the rectangular block.

Volume

= length ×breadth ×height

= 3.5 ×6 ×2.5

= 52.5cm³

Mass of the block

= 350(52.5)

= 18375g

Answer:

b rugby and basketball

Explanation:

basket ball was made by a white man

American Football originated from Soccer and Rugby. Hence, option (A) is correct.

American football, sometimes known as gridiron, is a team sport played by two teams of eleven players on a rectangular field with goalposts at each end. It is known simply as football in the United States and Canada. The defense, the team without the ball, seeks to halt the offense's movement and seize control of the ball for themselves.

The offence, the team in possession of the oval-shaped football, attempts to advance down the field by running with the ball or throwing it. The offence is awarded a new set of four downs to resume the drive if they successfully gain at least 10 yards in four downs or plays. If they fail, they forfeit the ball to the defense.

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Compression force can be measured with a force gage or load cell. 

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(c) The angular frequency of the oscillation is 7.885 rad/s.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

The period of the oscillation is the time taken to make one complete cycle.

T = 0.8 s

The frequency of the oscillation is calculated as follows;

[tex]f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz[/tex]

The angular frequency of the oscillation is calculated as follows;

[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s[/tex]

The force constant of the spring is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m[/tex]

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Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]

Now, we can solve for sigma using the other given condition:

[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]

c.

The angular displacement is the INTEGRAL of the angular velocity function.

[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]

[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]

[tex]\theta = 8.471 rad[/tex]

Convert this to rev:

[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]

Evaluate the improper integral:

[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]

Convert to rev:

[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]

Answer:

Explanation:

s = ½at²

s = ½(9.50)(29.3²) = 4,077.8275

s = 4080 m

when rounded to the three significant digits of the question numerals.

Answer:

F = 0 N

Explanation:

The problem does not talk about any acceleration. We can just assume it always moves at a constant speed so the resultant force will be 0 newtons. Of course you could say that it accelerates for a certain amount of time, as the wind blows, but the problem doesn't tell us when to calculate the resultant force (if during the acceleration or after) + we don't have the mass of the jet.

Answer:

H = Vy t - 1,2 g t^2            formula for height of ball after t sec

H = 10 - 1.3 = 8.7 m

Vy = 27 sin 47 = 19.7 m/s     vertical speed of ball

8.7 = 19.7 t - 9.8/2 t^2       height of ball after t sec

4.9 t^2 - 19.7 + 8.7 = 0       rearranging

[19.7 ± (388 - 170)^1/2] / 2 *4.9 = [19.7 ± 14.7] / 9.8 = .51 ,3.5 sec

.51 sec would be on the way up and 3.5 sec on the way down

Sx = 27 * cos 47 * 3.5 = 64.4 m around 200 ft seems reasonable

Explanation:

given solution

h=45m v^2=u^2+2gh

g=10m/s^2 v^2=0^2+2×10m/s^2×45m

vi=0 v^2=900m^2/s^2

Answer:

Explanation:

Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?

Obviously the sea is salt water so we can ignore  ρ = 1000 kg/m³

1025 kg/m³(d  m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa

d = 9.9450535...

d = 10 meters

That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa

so the robot would be a hair above its pressure limit before it even got in the water.

An object which has more electron than proton is negatively charged, otherwise positively charged. Every statement is false

Atom is the smallest unit of the matter consist of the positive charged nucleus and the electrons which moves around it. The atom can not be divided further.

The atom of a matter is made by three elements-

For the Protons and neutrons, lets check all the statement wheather they are true or false.

Thus, every statement is false as a object which has more electron than proton is negatively charged, otherwise positively charged.

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Answer:

Heat flow

Explanation:

Answer:

measure the area in which the rider will sit on, then take those measurements to a person who crafts saddles, tell them the exact measurements. then you wait a while until he/she is done then put the saddle on the horse.

Explanation:

Answer:

Explanation:

The shape of the horse's withers is also a determining factor in choosing a saddle. An adjustable hoop saddle allows you to choose with finesse the opening of the tree at the height of the withers, in order to best adapt to each horse. It is the size of the arch that determines the opening of the saddle.

English saddles are the main riding saddles used in equestrian disciplines around the world: Show jumping, endurance, require specific designs to meet specific needs. For example, dressage saddles are designed to bring the rider closer to the horse, allowing him to communicate very precisely with his horse.

Answer:

19.3m/s

Explanation:

Use third equation of motion

[tex]v^2-u^2=2gh[/tex]

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

[tex]v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s[/tex]

Answer:

1.108 ×  [tex]10^{-3}[/tex]J

Explanation:

v=21.0v

Q=52.8×  [tex]10^{-6}[/tex]

E=?

V=E/Q

E=v ×Q

 =21 ×52.8 ×[tex]10^{-6}[/tex]

 =1108.8   ×[tex]10^{-6}[/tex]

E= 1.108 ×  [tex]10^{-3}[/tex]J