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3 years ago

Jamie has a mass of 35kg what is her weight on earth in newtons

Physics

2 answers:

nevsk [136]3 years ago

8 0

Gravity on Earth = 10N/Kg
10N/Kg * 35Kg =350N

zvonat [6]3 years ago

3 0

On eart 1 kg is equal to 9.8 N/kg
So you just have to do 35•9,8= 343N
343 N is the answer.
And it is NOT 343 N/kg because the kilograms of 35 and the kilograms of 9,8 simplify (they cancel each other) so the remaining unit of measure is newton

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When a rigid body rotates about a fixed axis, all the points in the body have the same A. centripetal acceleration B. tangential

Leto [7]

Answer:

The angular acceleration is same at all the points in the body.

Option (D) is correct.

Explanation:

Given:

When a rigid body rotates about a fixed axis, all the points in the body have the same,

For finding which quantity is same we use pure rotational concept,

$v = \omega r$

Where $\omega =$ angular frequency, $r =$ radius of rigid body

When a rigid body rotates about a fixed axis angular velocity of all the points in the body are same.

But the tangential speed, tangential acceleration, linear displacement, and centripetal acceleration depend on the position of the points and hence they are not the same.

Therefore, the angular acceleration is same at all the points in the body.

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3 years ago

Does the universe have centre?

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3 years ago

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Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t

marshall27 [118]

Answer:

x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

the initial position of car a is zero

x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

x = x_{ob} + v_{ob} t - ½ a_b t²

car B's starting position is 30 m

x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

v_{ob} = 23.4 km / h = 6.5 m / s

4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

0.2 t² - 2.5 t - 30 = 0

t² - 12.5 t - 150 = 0

we solve the quadratic equation

t = $\frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150} }{2}$

t = $\frac{12.5 \ \pm 27.5}{2}$

t₁ = 20 s

t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

x = 4 t + ½ 0.8 t²

x = 4 20 + ½ 0.8 20²

x = 240 m

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3 years ago

An old clock has a spring that must be wound to make the clock hands move. Which statement describes the energy of the spring an

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3 years ago

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In a glider stunt at an air show, a towing airplane (motorized plane pulling the gliders) takes off from a level runway with two

irakobra [83]

Answer:

minimum length of runway is needed for take off 243.16 m

Explanation:

Given the data in the question;

mass of glider = 700 kg

Resisting force = 3700 N one one glider

Total resisting force on both glider = 2 × 3700 N = 7400 N

maximum allowed tension = 12000 N

from the image below, as we consider both gliders as a system

Equation force in x-direction

2ma = T -f

a = T-f / 2m

we substitute

a = (12000 - 7400 ) / (2 × 700 )

a = 4600/1400

a = 3.29 m/s²

Now, let Vf be the final speed and Ui = 0 ( as starts from rest )

Vf² = Ui² + 2as

solve for s

Vf² = 0 + 2as

2as = Vf²

s = Vf² / 2a

given that take of speed for the gliders and the plane is 40 m/s

we substitute

s = (40)² / 2×3.29

s = 1600 / 6.58

s = 243.16 m

Therefore, minimum length of runway is needed for take off 243.16 m

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3 years ago