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2 years ago

PLEASE HELP

Physics

2 answers:

Vlad1618 [11]2 years ago

8 0

C) end Q attracts both ends X and Y.

stepladder [879]2 years ago

3 0

Answer:

here's the answer to your question hope it helps you

Explanation:

b. is the answer

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How much force does a soccer goalie

Tems11 [23]

Answer:

2 N

Explanation:

From the question, it's given that

Mass m = 0.2 kg

Acceleration a = 10 m/s^2

The force a soccer goalie experience when stopping a ball will be equal to the force at which the ball is being kicked. This is

F = ma

Substitute all the parameters into the formula

F = 0.2 × 10

F = 2 Newton.

3 0

3 years ago

Please Help

musickatia [10]

The answer is C because <span>this movement is caused by the heat in Earth and creates </span>convection currents. <span>Convection currents in the asthenosphere cause movement of Earth's tectonic plates. </span>

4 0

3 years ago

Read 2 more answers

Many biological systems are well-described by the laws of statistical physics. A simple yet often powerful approach is to think

GuDViN [60]

Answer:

z1/z2

Explanation:

we have no quantum effects therefore we can make use of Maxwell Boltzmann distribution in the description of this system.

using the boltzman distribution the probability of finding a particle in energy state

$P_{ei} = \frac{gie^{-ei/kol} }{z}$

we have

gi to be degeneration of the ith state

ei to be energy of ith state

$z=e^{-ei/kbt}$ summation

$P_{ope} = \frac{e^{-ei/kBt} }{z} = \frac{Z_{1} }{Z}$

We have R to be equal to

$\frac{P_{ope} }{P_{Close} } = \frac{Z1}{Z2}$

8 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

8 0

3 years ago