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Oxana [17]
3 years ago
12

A 70 kg man is walking at a speed of 2 m/s. What is his Kinetic Energy?

Physics
2 answers:
Ilya [14]3 years ago
8 0

Kinetic energy = 1/2 (mass) (speed squared)

KE = 1/2 (70kg) (2m/s)^2

KE = (35 kg) (4 m^2/s^2)

KE = (140) kg-m^2/s^2

That's 140 Newton-meter

or

140 Joules

den301095 [7]3 years ago
5 0
140 Jules. Is the correct answer
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Briefly describe the formation of the planets from the solar nebula
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6 0
3 years ago
A maple tree seed fell 180 centimeters straight toward the ground at a constant velocity. It moved that distance in 1.5seconds.
denis23 [38]

Answer: 1.2 m/s

Explanation:

Velocity V is defined as the variation of position of an object or body in time. So, if we know the distance the seed traveled and the time, we can calculate its velocity:

V=\frac{d}{t}

Where:

d=180 cm \frac{1 m}{100 cm}=1.8 m is the distance the maple seed traveled

t=1.5 s is the time

Then:

V=\frac{1.8 m}{1.5 s}

V=1.2 m/s This is the seed's velocity

7 0
3 years ago
How close does the proton get to the line of charge?
loris [4]

Answer:

12 cm

Explanation:

3 0
2 years ago
A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and
ycow [4]

Answer:

Acceleration a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

Explanation:

For the truck to accelerate without losing its load.

Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.

Fa ≤ F(friction)

But;

Fa = mass × acceleration

Fa = ma

ma ≤ F(friction)

a ≤ (F(friction))/m ......1

Given;

Fa = mass × acceleration

Fa = ma

mass m = 800 kg

F(friction) = 2400 N

Substituting the given values into equation 1;

a ≤ F(friction)/m

a ≤ 2400N/800kg

a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

4 0
3 years ago
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
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