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3 years ago

Please help, I am stumped.

Physics

2 answers:

ehidna [41]3 years ago

6 0

30 is the correct answer your looking for

Ivahew [28]3 years ago

4 0

Answer: A) 30 grams

Explanation: According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants.

mass of reactants= mass of antacid + mass of water= 30 g+ 50 g= 80 g

Given : Mass of product = 50 g

But mass of products should be equal to mass of reactants = 80 g

Thus mass of another product= (80-50)g = 30 g.

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An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution

givi [52]

With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
= (15π rad/s)*(1 rev/2π rad)
= 7.5 rev/s

Answer: 7.5 revolutions per second.

7 0

4 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity

umka21 [38]

Answer: $29.85\ W/m^2$

Explanation:

Given

Power $P=60\ W$

Distance from the light source $r=0.4\ m$

Intensity is given by

$I=\dfrac{P}{4\pi r^2}$

Inserting values

$\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2$

3 0

3 years ago

Read 2 more answers

If you kick a ball and the ball accelerates through the air, the forces involved must be

wariber [46]

It would be kinetic and potential energy, potential from the force of the kick, and kinetic from energy exerted.

7 0

3 years ago

A 20 liter cylinder of helium at a pressure of 150 atm and a temperature of 27ÁC is used to fill a balloon at 1.00 atm and 37ÁC.

djyliett [7]

<u>Answer</u>

D) 3100 Liters

<u>Explanation</u>

To get the volume if the balloon you need to use the combined equation of the low of gases.

P₁V₁/T₁ = P₂V₂/T₂

(20×150)/(27+273) = (1×V₂)/(37+273)

3000/300 = V₂/310

10 = V₂/310

V₂ = 10 × 310

= 3100 Liters

7 0

3 years ago

Read 2 more answers