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2 years ago

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What is know as law of inertia?

1 answer:

schepotkina [342]2 years ago

5 0

Answer:

<em><u>mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>brianliest</u></em><em><u> </u></em><em><u>plz</u></em>

Explanation:

Law of inertia, also called Newton's first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
Law of Inertia states that a body in a state of rest or uniform motion remains in the same state until and unless an external force acts on it.
A body continues to be in its state of rest or in uniform motion along a straight line unless an external force is applied on it. This law is also called law of inertia.

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What is peer review?...

Bond [772]

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2 years ago

Read 2 more answers

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

3 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

2 years ago

An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam

Amanda [17]

Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V = the volume of the sample.
Then
W = ρVg
or
V = W/(ρg)
= (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
= 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³

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2 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth.

Maksim231197 [3]

Answer:

Explanation:

We shall apply law of conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

$\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0$

$\frac{-GMm}{R} +\frac{1}{2}\times \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}$

$\frac{0.877GMm}{R} =\frac{-GMm}{R+h}$

h = .14 R

b )

$\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0$

$\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}$

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

= GMm / R + 1/2 m (2GM/R)

= 0

5 0

3 years ago