All categories

6 days ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

s sitting at the bottom of the incline. Assume the first block comes to rest after the collision. The second box travels over a rough surface (µk = 0.5) for 2m. At the end of the rough surface there is a spring of spring constant 250N/m. The spring is on a frictionless surface. By how much would the second block compress the spring?

Physics

You might be interested in

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

Maru [3345]

Answer:

The excess charge is $Q = 3.5 *10^{-7} \ C$

Explanation:

According to the question, we are informed that

The diameter is $d = 45 \ cm = 0.45 \ m$

The potential of the surface is $V = 14 \ kV = 14 *10^{3} \ V$

The radius of the sphere is

$r = \frac{d}{2}$

by plugging in given values

$r = \frac{0.45}{2}$

$r = 0.225 \ m$

The potential at the surface is mathematically expressed as

$V = \frac{k * Q }{r }$

Where k is Coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

Based on the question stating there are no other charges, Q represents the excess charge

Therefore

$Q = \frac{V* r}{ k}$

inserting the numerical values

$Q = \frac{14 *10^{3} 0.225}{ 9*10^9}$

$Q = 3.5 *10^{-7} \ C$

7 0

2 months ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

serg [3582]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We utilize the formula

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the angular momentum magnitude of the person 2 meters prior to jumping onto the merry-go-round?

The equation we apply is

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the angular momentum of the person just before she hops onto the merry-go-round?

We utilize the formula

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular velocity of the merry-go-round after the individual jumps on?

We can apply the Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒ I₀*ω₀ = (I₀ + m*R²)*ω₁

At this point, we can determine ω₁

⇒ ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒ ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒ ω₁ = 0.769 rad/s

5) Once the merry-go-round moves at this new angular speed, what force must the person exert to hold on?

We must calculate the centripetal force as follows

Fc = m*ω²*R

⇒ Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) When the person is halfway around, they choose to simply release their hold on the merry-go-round to exit the ride.

What is the linear speed of the person the moment they exit the merry-go-round?

we can apply the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular velocity of the merry-go-round after the individual releases their hold?

ω₀ = 1.53 rad/s

It returns to its original angular speed

8 0

2 months ago

Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights

inna [3103]

275 kPa Explanation: Here the mass of the gas equals m=1.5 kg with an initial volume of V₁=0.04 m³ and an initial pressure P₁=550 kPa. As provided, the final volume is double the original volume, making V₂ equal to 2 V₁. Since the temperature remains constant, T₁=T₂=T. By substituting the values into the equation... results in final pressure being P₂=275 kPa.

3 0

2 months ago

How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov

Keith_Richards [3271]

Answer:

x = v₀ cos θ t, y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.

Now, we will apply trigonometry to determine the initial velocity components.

sin θ = $v_{oy}$ / v₀

cos θ = v₀ₓ / v₀

v_{y} = v_{oy} sin θ

v₀ₓ = v₀ cos θ

Next, let's formulate the equations of motion.

X axis

x = v₀ₓ t

x = v₀ cos θ t

vₓ = v₀ cos θ

Y axis

y = y₀ + $v_{oy}$ t - ½ g t2

y = y₀ + v₀ sin θ t - ½ g t2

v_{y} = v₀ - g t

v_{y} = v₀ sin θ - gt

$v_{y}^{2}$ = v_{oy}^2 sin² θ - 2 g y

It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components

7 0

2 months ago