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ExtremeBDS [4]
3 years ago
14

How fast does a cats hair grow back?

Physics
2 answers:
Lady_Fox [76]3 years ago
6 0
It depends on the type of cat....
FromTheMoon [43]3 years ago
4 0
Most cats will grow their coat back within 8-12 weeks after shaving. In the case of my cats, it didn't take nearly long enough. Here they are sporting a 'lion cut' they had this summer in an attempt to cut down on the tumbleweeds of fur around the house.
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The lever shown above can be used to move the
vladimir1956 [14]

Answer:

C your answer would be C

Explanation:

It should be right

4 0
3 years ago
If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)
taurus [48]
Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20    Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

In calories,  4.2 J = 1 Calorie
=  8 400 000 / 4.2   = 200 000

Q = 200 000 Calories
4 0
3 years ago
A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th
Marat540 [252]

Answer:

a)  19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

b)

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

t = 10 sec

5 0
3 years ago
A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?
prohojiy [21]

Answer:

<h3>62.5N</h3>

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the  force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

8 0
3 years ago
A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and a
Elodia [21]

Answer:

Elastic potential energy, E = 3.26 J

Explanation:

It is given that,

Force constant of the spring, k = 5.2 N/m

Relaxed length of the spring, X = 2.45 m

When the mass is attached to the end of the spring, the vertical length of the spring is, x' = 3.57 m

To find,

The elastic potential energy stored in the spring.

Solution,

The extension in the length of the spring is given by :

x=x'-X

x=3.57\ m-2.45\ m

x = 1.12 m

The elastic potential energy of the spring is given by :

E=\dfrac{1}{2}kx^2

E=\dfrac{1}{2}\times 5.2\times (1.12)^2

E = 3.26 J

So, the elastic potential energy stored in the spring is 3.26 joules.

6 0
3 years ago
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