What is the average momentum of Cart 1 during the entire time shown before
kg-m
the collision, in units .?
$
Note that the collision appears to take place somewhere between 0.5 s and 0.6
s (perhaps at 0.6 s exactly, but we cannot be sure), so we can safely say that
data up to and including 0.5 s is "before the collision."

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

Answer:

Heat flow

Explanation:

Answer:

*1) 100 Joule energy

*2) 101.2 m/s

Explanation:

*1) 1J = 1w

100J = 100w

*2) A 70-kg person will have to run at a speed of 101.2 m/s to have that amount of kinetic energy.

Answer:

Explanation:

70.0(5.00)(9.81) = 3,433.5 = 3430 N

To solve this we must be knowing each and every concept related to  centripetal force and its calculations. Therefore, the centripetal force acting on her is 3430 N.

The term centripetal relates to a propensity to gravitate toward the center. Centripetal refers to moving in the direction of the center. The force that maintains an item moving in a circular direction and helps it stay on the path is referred to as centripetal force.

Furthermore, centrifugal force is indeed the tendency of things to deviate from a circular route and fly in a straight line. People frequently confuse centripetal force with centrifugal force.

Mathematically,

F = m a

  = 70 acceleration

  = 70 × 5 × 9.81

 = 3430 N

Therefore, the centripetal force acting on her is 3430 N.

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Answer:

a

Explanation:

i took the test 100%

Answer:

what will happen if i will answer ur questions?

Explanation:

is there gonna be a bad thing or a good thing

Answer:

lithosphere

Explanation:

hope this helps you!!

Answer:

Explanation:

Conservation of momentum

In the x direction

2(15) + 2(-10) = 2(-5) + 2(vBfx)

vBfx = 10 m/s

In the y direction

2(30) + 2(5) = 2(20) + 2(vBfy)

vBfy = 15 m/s

vBf = 10i + 15j

KEi = ½(2)(15² + 30²) + ½(2)(-10² + 5²) = 1250 J

KEf =  ½(2)(-5² + 20²) + ½(2)(10² + 15²) = 750 J

KEf - KEi = 750 - 1250 = -500 J

Answer: the current increases with each 3 volt and 9 volt. The relationship between resistance and current in a circuit is that the greater the resistance the less the current and the greater the current the less the resistance is. yayayay I could answer this I big brain :)

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

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

Answer:

Explanation:

conservation of momentum

m(u) + M(0) = (m + M)v

u = (m + M)v/m

u = (0.250 + 1.35)(11.9) / 0.250

u = 76.16

u = 76.2 m/s

That's a fairly massive, and slow, bullet.

Answer:

Adding salt to the water increases the density of the solution because the salt increases the mass without changing the volume very much.

Explanation: the explanation is in a file

Answer:

Explanation: 6000z

The muzzle speed of the bullet before the collision is 415.3 m/s.

The given parameters:

Apply the principle of work-energy theorem to determine the final velocity of the block-bullet system;

[tex]F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;

[tex]m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s[/tex]

Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.

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Hi there!

To solve, we must begin by calculating the total WORK done on the object.

W = F · d (Force · displacement)

Plug in the given values. Remember to convert km to m:

1 km = 1000 m

0.6 km = 600 m

W = 60 · 600 = 36000 J

Now, we can solve for power:

P = W/t

Convert minutes to seconds:

1 min = 60 sec

20 min = 1200 sec

P = 36000/1200 = 30 W ⇒ Choice D.

Answer:

Increasing the number of trials reduces the impact of any one imprecise measurement. … To increase the number of attempts, you can find an average result for the experiment, as well as find and discrepancies as human error if you perform an experiment several times.

Explanation:

hope it helps :)

Answer:

It is because the increase in the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value.

Answer:

Explanation:

a) F = ma

a = F/m

a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s

b) t = v/a

t = 200 / 7.2 x 10⁻⁶

t = 2.8 x 10⁷ s       about 10½ months

c) v² = u² + 2as

s = (v² - u²) / 2a

s = (200² - 0²) / (2( 7.2 x 10⁻⁶))

s = 2.8 x 10⁹ m    nearly 7 times around the earth

And all this assumes NO FRICTION.

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:

Explanation:

The word "sheet" implies that the copper is quite thin.

Copper is also a very good conductor of heat.

Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.

Answer:

i cant see

Explanation:

but im smart

[tex]\\ \sf\Rrightarrow T=F[/tex]

[tex]\\ \sf\Rrightarrow T=ma[/tex]

[tex]\\ \sf\Rrightarrow T=100(2)[/tex]

[tex]\\ \sf\Rrightarrow T=200N[/tex]

All the three concurrent forces acting on a body will not produce equilibrium.

The given parameters:

1. 1 N, 3 N and 5 N

2. 2N, 2N and 2 N

3. 3N, 4N and 5 N

4. 4N, 4N and 5 N

Concurrent forces lie on the same plane and their line of action pass through a common point.

A body under concurrent forces is in equilibrium if the resultant of the forces on the body is zero.

[tex]\Sigma F = 0\\\\F_1 + F_2 + F_3 = 0\\\\F_1 + F_ 2 = - F_3[/tex]

where;

[tex]F_3[/tex] is the equilibrant force

First set of concurrent forces;

[tex]1 \ N \ + \ 3\ N = 4 \ N\\\\F_ 3 = 5 \ N\\\\5 \ N > 4 \ N[/tex]

Second set of concurrent forces;

[tex]2 \ N \ + \ 2 \ N = 4 \ N\\\\F_ 3 = 2 \ N\\\\4 \ N > 2 \ N[/tex]

Third set of concurrent forces;

[tex]3 \ N \ + \ 4 \ N = 7 \ N\\\\F_ 3 = 5 \ N\\\\7 \ N > 5 \ N[/tex]

Fourth set of concurrent forces;

[tex]4 \ N \ + \ 4 \ N = 8 \ N\\\\F_ 3 = 5 \ N\\\\8 \ N > 5 \ N[/tex]

Thus, we can conclude that all the three concurrent forces acting on a body will not produce equilibrium.

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Answer:

Explanation:

F = mv²/R

F = 11.0(3.50²)/0.600 = 225 N

Answer:

Explanation:

Average velocity in the 24 m interval is 24 / 4 = 6 m/s

Average velocity in the 64 m interval is 64 / 4 = 16 m/s

There is a 4 second interval between the two points where average velocity equals actual velocity

a = Δv/t = (vf - vi) / t = (16 - 6) / 4 = 2.5 m/s²

s = v₀t + ½at²

24 = v₀(4) + ½(2.5)4²

4v₀ = 24 - 20

v₀ = 1 m/s

Not asked for but the velocity at the end of the first segment and beginning of the second segment is 11 m/s and final velocity is 21 m/s

Answer:

The removal of heat energy slows the speed of particles

Explanation:

When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]

[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]

[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]

[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]

[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]

[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]

[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

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The angular velocity of the wheel depends on the mass, radius and the

mode of rotation of the wheel (with or without slipping).

Reasons:

The given parameters are;

Radius of the wheel, r = 2.0 m

Height of the incline, h = 8.0 m

Required:

Angular velocity of the wheel at the bottom of the incline.

Solution:

The potential energy of the wheel at the top of the hill, P.E. = m·g·h

[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]

Where;

v = The translational velocity of the wheel = ω·r

I = The moment of inertia of the wheel = m·r²

Therefore'

[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]

At the bottom of the hill, the potential energy is converted to kinetic energy

Therefore;

P.E. = Sum of K.E.

m·g·h = m·r²·ω²

g·h = r²·ω²

[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]

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Crazy Wally Ok Ok ok hhahahaha