Look at the circuit given below. It consists of a cell, a bulb with two terminals X, Y and wires. P, Q, R and S are positions marked. What is the direction of the flow of current? a) PQXYRS b) SRYXQP c) SPQXYR d) PSRYXQ

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:

PE = mgh

g = PE/mh

g = 15.0 / (0.200(0.500))

g = 150 m/s²

This is one strong astronaut if he can work in an environment where gravity is more than 15 times stronger than on earth.

Answer:

Heat flow

Explanation:

Answer:

i cant see

Explanation:

but im smart

Answer:

Explanation:

30 m/s • 3 hr •3600 s/hr / 1000 m/km = 324 km west

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

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 :)

Crazy Wally Ok Ok ok hhahahaha

Answer:

n

Explanation:

TWO outcomes can be predicted the students will find:

When a constant mass of gas is cooled, its volume falls, and when the temperature is raised, its volume grows. The volume of the gas rises by 1/273 of its initial volume at 0 °C for every degree of temperature rise.

In layman's words, the volume of a fixed mass of gas is exactly proportional to temperature at constant pressure.

The teacher took two latex balloons and blew them up with helium gas to the same size. As she placed Balloon A in a -15° C freezer, its temperature decreases and that's why, the size of balloon A becomes smaller. Again she  placed Balloon B in the sun on 30° C day, its temperature increases and that's why, the size of balloon B becomes larger.

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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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Answer:

inertia   .

because yes

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:

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:

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:

a

Explanation:

i took the test 100%

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:

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:

what will happen if i will answer ur questions?

Explanation:

is there gonna be a bad thing or a good thing

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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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.

Answer:

[tex]\frac{a_{d}}{a_{i}} = \frac{(1 -mu)}{mu}[/tex]

= (1 - μ)/μ

Explanation:

Always draw a diagram!

Up the incline:

[tex]Fr_{max}[/tex] = maximum friction

[tex]Fr_{max}[/tex] = μk

k = R = mg.cos(45) = mg.sin(45)

Resolution of forces parallel to the slope:

F (Fp in the diagram) = force of propulsion

g = gravity

[tex]F - Fr_{max} = ma_{i}[/tex]

[tex]F -[/tex] μ.mg.cos(45) [tex]= ma_{i}[/tex]

Down the decline:

Resolution of forces:

[tex]mg.sin(45) - Fr_{max} = ma_{d}[/tex]

[tex]mg.sin(45) -[/tex] μ.mg.cos(45) [tex]= ma_{d}[/tex]

Then, find the ratio:

[tex]\frac{ma_{d}}{ma_{i}} = \frac{mg.sin(45) - mu.mg.cos(45)}{-F + mu.mg.cos(45)} \\\\ \frac{a_{d}}{a_{i}} = \frac{k - k.mu}{-F + k.mu} \\\\ = \frac{k(1 -mu)}{-F + k.mu}[/tex]

Potentially, there is no need to consider F in this situation, in which case:

[tex]\frac{a_{d}}{a_{i}} = \frac{k(1 -mu)}{k.mu} \\\\ = \frac{(1 -mu)}{mu}[/tex]

= (1 - μ)/μ

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:

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:

lithosphere

Explanation:

hope this helps you!!

Answer:

Explanation:

F = mv²/R

F = 11.0(3.50²)/0.600 = 225 N

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