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vfiekz [6]
2 years ago
4

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in time t, th

rough what angle did it rotate in the time ½t?
a. 4θ
b. ¼θ
c. ½θ
d. 2θ
e. θ
Physics
1 answer:
Blizzard [7]2 years ago
4 0

Answer:

Angular displacement will be \frac{1}{4}\Theta

So option (b) will be the correct option

Explanation:

We have given that firstly object is at rest

So \omega _i=0rad/sec

From law of motion we know that angular displacement is given by

\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2

Now angular displacement by the object in \frac{t}{2}sec

\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta

So option (b) will be the correct option

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What is NOT one of the three primary resources that families have to reach financial goals?
Elodia [21]
What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
8 0
2 years ago
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block
melomori [17]

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

5 0
2 years ago
Question
nataly862011 [7]

The density of a material is constant and it is given by the ratio of the mass to the volume of the material

The mass of the liquid and the full bottle ae

The mass of the liquid is <u>450 g</u>

The mass of the filled bottle is <u>530 grams</u>

<u></u>

The reason the above values are correct are as follows:

The given parameters are;

Volume of the bottle, V = Half litre

Mass of the bottle, m_b = 80 g

The volume of liquid in the bottle when filled, V = 500 cm³

The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³

a. Required:

To calculate the mass of oil in the bottle

Solution:

The volume of oil in the bottle when the bottle is filled, V = 500 cm³

Density, \ \rho = \dfrac{Mass}{Volume}  = \dfrac{m}{V}

The mass of the liquid, m = ρ × V

∴ m = 0.9 g/cm³ × 500 cm³ = 450 g

The mass of the liquid, m = <u>450 g</u>

<u></u>

b. The mass of the oil in the bottle, m = grams

The mass of the full bottle, m_{filled} = m + m_b

∴ m_{filled} = 450 g + 80 g = 530 g

The mass of the full bottle, m_{filled} = <u>530 grams</u>

Learn more about density here:

brainly.com/question/18110802

4 0
2 years ago
A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin
aleksandr82 [10.1K]

Answer:

Explanation:

Given

volume of sample V_1=90\ mL

Temperature T_1=298\ K

Pressure P_1=702\ mm\ of\ Hg

for different conditions

Temperature T_2=273\ K

Pressure P_2=760\ mm\ of\ Hg

suppose V_2 is the volume of sample

Using ideal gas equation

PV=nRT

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

\frac{702\times 90}{298}=\frac{760\times V_2}{273}

V_2=76.15\ mL

             

8 0
2 years ago
the steel bed of a suspension bridge is 200m long at 20 C. If the extremes of temperature to which it might be exposed are -30 C
raketka [301]

Answer:

The steel bed will contract by  0.13 m, and expand by 0.052 m

Explanation:

For contraction,

α = ΔL/(LΔθ)..................... Equation 1

Where α = Linear expansivity of  steel, ΔL = decrease in length/ Increase in length, L = Original length, Δθ = Change in temperature

make ΔL the subject of the equation

ΔL = α(LΔθ)................. Equation 2

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = -30-20 = -50 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(-50)

ΔL = -0.13 m

Similarly, For expansion,

Using equation 2

ΔL =  α(LΔθ)

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = 40-20 = 20  °C

Substitute into equation 2

ΔL =  13×10⁻⁶(200)(20)

ΔL = 0.052 m.

Hence the steel bed will contract by  0.13 m, and expand by 0.052 m

3 0
2 years ago
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