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kirill [66]
3 years ago
13

A cat chases a mouse across a 0.63 m high

Physics
1 answer:
Ber [7]3 years ago
3 0

Answer:

6.7 m/s

Explanation:

In the vertical direction:

y₀ = 0.63 m

y = 0 m

v₀ᵧ = 0 m/s

aᵧ = -9.81 m/s²

In the horizontal direction:

x₀ = 0 m

x = 2.4 m

aₓ = 0 m/s²

Find: v₀ₓ

First, find the time:

y = y₀ + v₀ᵧ t + ½ aᵧt²

0 = 0.63 + (0) t + ½ (-9.81) t²

t = 0.358

Now, find the velocity:

x = x₀ + v₀ₓ t + ½ aₓt²

2.4 = 0 + v₀ₓ (0.358) + ½ (0) (0.358)²

v₀ₓ = 6.70

Rounded to two significant figures, the cat's velocity when it slides off the table is 6.7 m/s.

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A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s
Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

3 0
3 years ago
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t
Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

4 0
4 years ago
Latitude and longitude picture.​
TiliK225 [7]

Answer:

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Explanation:

7 0
3 years ago
Another athlete using a different spring exerts an average force of 400 N to enable her
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Answer:84Nm

Explanation:

force=400N

Distance=0.210m

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Workdone=84Nm

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What is the mass of a neutron?
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Answer:

1

Explanation:

8 0
3 years ago
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