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jeka57 [31]
3 years ago
8

A 65kg person has a weight of

Physics
2 answers:
sergij07 [2.7K]3 years ago
7 0

If you are looking for the metric term for weight, it would be weighed in Newtons, and that would be 637 Newtons.

If you are looking for pounds, it would be 143 lbs.

yawa3891 [41]3 years ago
6 0

Answer:

So on Earth, with its 9.8 metre per second gravity, a person with a mass of 65 kilograms actually weighs 637 Newtons. Acceleration due to gravity on the Moon is about a sixth of earth, 1.622 meters per second. As a result a person with a mass of 65 kilograms on the Moon would actually weigh 105.43 Newtons.

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A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end
Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0
3 years ago
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

where E = electric field intensity

5 0
3 years ago
Which type of electromagnetic radiation carries the most energy and has the highest frequency?
jek_recluse [69]

Answer:

Gamma rays

Explanation:

Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.

8 0
4 years ago
Read 2 more answers
A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th
Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

now we have

W_{gravity} = -mg(Lsin\theta)

W_{friction} = -\mu mgcos(\theta) L

so from above equation

Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

so from above equation work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

8 0
4 years ago
A 190 N child is in a swing that is attached to ropes 2.10 m long. Find the gravitational potential energy of the child-Earth sy
Alekssandra [29.7K]

Answer:

a) Gravitational potential energy = 399 J

b) Gravitational potential energy = 66.5 J

c) Gravitational potential energy = 0 J

Explanation:

Hi there!

Please, see the attached figure for a better understanding of the problem.

a) When the ropes are horizontal, the height of the child, relative to the child's lowest position, is 2.10 m (see figure).

The gravitational potential energy is calculated as follows:

PE = mgh

Where:

PE = potential energy.

mg = weight of the child

h = height.

Then when the ropes are horizontal, the potential energy will be:

PE = 190 N · 2.10 m = 399 J

b) When the ropes make a 34.0° with the vertical, the height of the child is 2.10 m minus x (see figure). To find x, we can use trigonometry of right triangles:

cos angle = adjacent side / hypotenuse

cos 34.0° = x / 2.10 m

x = 2.10 m · cos 34.0° = 1.75 m

Then, the height of the child relative to the lowest position is

(2.10 m - 1.75 m) = 0.35 m

Therefore, the gravitational potential energy will be:

PE = 190 N · 0.35 m

PE = 66.5 J

c) When the child is at the bottom of the circular arc the height is zero (the child is at the lowest position), then, the gravitational potential energy will be zero.

6 0
3 years ago
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