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3 years ago

What did the hand say to the face?

Physics

2 answers:

77julia77 [94]3 years ago

7 0

What did it say did it say “slap

RSB [31]3 years ago

6 0

Answer:

im not sure what did it say

Explanation:

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

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Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw

jarptica [38.1K]

Answer:

a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

θ = 90 cos 90 = 0

W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

θ = 180 cos 190 = -1

W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

m g sin θ L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀= + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

ΔU = -U₀

the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

W_Total = W_weight - W_roce

as the body moves down

W_Total> 0

Therefore the total work is positive

3 0

3 years ago

car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 k

Mice21 [21]

Answer:

32000 N

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Mass (m) of car = 400 Kg

Force (F) =?

Next, we shall determine the acceleration of the the car. This can be obtained as follow:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Acceleration (a) =?

v² = u² + 2as

0² = 40² + (2 × a × 10)

0 = 1600 + 20a

Collect like terms

0 – 1600 = 20a

–1600 = 20a

Divide both side by –1600

a = –1600 / 20

a = –80 m/s²

The negative sign indicate that the car is decelerating i.e coming to rest.

Finally, we shall determine the force needed to stop the car. This can be obtained as follow:

Mass (m) of car = 400 Kg

Acceleration (a) = –80 m/s²

Force (F) =?

F = ma

F = 400 × –80

F = – 32000 N

NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.

7 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

3 years ago

look at the circuit in the figure. find the current, voltage, and power in each resistor. please list answer

sashaice [31]

Answer:

fhcjctfkbraf gdovtckcrhha6g

6 0

3 years ago

Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angul

vredina [299]

Answer:

D) True. the protostar rotates more quickly.

Explanation:

If the system is isolated, the angular momentum must be retained.

Initial

L₀ = I w₀

Final

$L_{f}$ = $I_{f}$ $w_{f}$

L₀ = $L_{f}$

I w₀ = $I_{f}$ $w_{f}$

$w_{f}$ = I / $I_{f}$ w₀

In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase

Let's examine the answers

A) False. The opposite happens

B) False. Speed changes

C) False. For this there must be an external force, which does not exist

D) True. You agree with the above

5 0

3 years ago