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3 years ago

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Knowing that a 0.90-kg object weighs 9.0 N , find the acceleration of a 0.90-kg stone in free fall . Express your answer to two

significant figures and include the appropriate units.

1 answer:

sasho [114]3 years ago

7 0

Answer:

The value of acceleration of a stone in free fall is $10\ m/s^2$ .

Explanation:

Given that,

Mass of the object, m = 0.9 kg

Weight of the object, W = 9 N

We need to find the acceleration of a stone in free fall. The weight of an object is given by :

W = mg

g is the acceleration of the stone in free fall

So, $g=\dfrac{W}{m}$

$g=\dfrac{9\ N}{0.9\ kg}$

$g=10\ m/s^2$

So, the value of acceleration of a stone in free fall is $10\ m/s^2$ . Hence, this is the required solution.

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How to find a planet’s gravitational field strength using its radius?

grin007 [14]

The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

$g = \frac{F}{m}$

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And $F =\frac{GMm}{r^{2} }$

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

$g = 0.977 \times 10^1= 9.77\ N$

So, the gravitational field strength is approximately equal to 10 N.

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3 years ago

Examine the scenario. Object A has 5 protons and 5 electrons. Object B has 5 protons and 7 electrons. Which option most accurate

Travka [436]

<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
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4 0

3 years ago

A rotating merry go-round makes one complete revolution in 4.2 s. (a) what is the linear speed of a child seated 1.3 m from the

gregori [183]

The radius, r, of the child from the center of the wheel is
r = 1.3 m

The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s

The linear speed of the child is the tangential velocity, given by
v = rω
= (1.3 m)*(1.496 rad/s)
= 1.945 m/s

Answer: 1.95 m/s (nearest hundredth)

6 0

3 years ago

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1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of

VLD [36.1K]

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

6 0

3 years ago

4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle ( $\Phi$ ) = 20°

Pinion speed ( $n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

= $\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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3 years ago

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