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Lera25 [3.4K]
3 years ago
13

What is the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s?

Physics
2 answers:
Alik [6]3 years ago
3 0
1/2mv^2

1/2x12x10^2=600J

The kinetic energy is 600J
Katena32 [7]3 years ago
3 0

Answer:

The kinetic energy of the object is 600 J

Explanation:

It is given that,

Mass of the object, m = 12 kg

Velocity of the object, v = 10 m/s

Kinetic energy of the object is given by :

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

E=\dfrac{1}{2}\times 12\times (10)^2

E = 600 J

So, the kinetic energy of the object is 600 J. Hence, this is the required solution.

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A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and
N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t    where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec =  1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r  in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0
3 years ago
Read 2 more answers
To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m
Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

T_{2} = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

8 0
3 years ago
If it takes 42 minutes to load 3 1/2 trucks, how many minutes will it take to load 6 1/2 trucks?
Kruka [31]
If it takes 42 minutes to load 3 1/2 trucks, then it takes 42 / 3.5 = 12 minutes to load a single truck. Multiplying the number of minutes per truck by 6.5 yields the time it will take to load. In this case 12 x 6.5 = 78 minutes would be required to load 6 1/2 trucks. The formula for the time taken is t = 12n, where t is the time in minutes and n is the number of trucks. 
6 0
3 years ago
A charged capacitor is connected to an ideal inductor. At time t = 0, the charge on the capacitor is equal to 6.00 μC. At time t
Len [333]

Answer:

4.71\times 10^{-3}A

Explanation:

Q_{max} = Maximum charge stored by capacitor = 6 μC = 6 x 10⁻⁶ C

t  = time taken for charge on the capacitor to become zero = 2 ms = 2 x 10⁻³ s

Time period is given as

T = 4t

T = 4(2\times 10^{-3})

T = 8\times 10^{-3} s

Angular frequency is given as

w = \frac{2\pi }{T}

w = \frac{2(3.14) }{8\times 10^{-3}}

w =785 rad/s

Charge at any time is given as

Q(t) = Q_{max}Coswt

Taking derivative both side relative to "t"

\frac{\mathrm{d}Q(t) }{\mathrm{d} t} = \frac{\mathrm{d}(Q_{max}Coswt) }{\mathrm{d} t}

i(t)= -Q_{max} w Sinwt

Amplitude of the current is given as

i_{max}= Q_{max} w

i_{max}= (6\times 10^{-6}) (785)

i_{max}= 4.71\times 10^{-3}A

8 0
3 years ago
How are hot spots used to track plate motion
DanielleElmas [232]
When plates move they sometimes open small cracks that release heat from under earths plate, coming from the mantle or even the core at times, this heat allows scientists to determine where the plates are moving from and where they are moving to.
5 0
3 years ago
Read 2 more answers
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