Give you point but please i need help in physcis

san4es73 [151]

Your answer would be a, it is a force.

6 0

3 years ago

A 97.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.67 r

Yuliya22 [10]

Answer:

w = 1.14 rad / s

Explanation:

This is an angular momentum exercise. Let's define a system formed by the three bodies, the platform, the bananas and the monkey, in such a way that the torques during the collision have been internal and the angular momentum is preserved.

Initial instant. The platform alone

L₀ = I w₀

Final moment. When the bananas are on the shelf

we approximate the bananas as a point load and the distance is indicated

x = 0.45m

L_f = (m x² + I ) w₁

angular momentum is conserved

L₀ = L_f

I w₀ = (m x² + I) w₁

w₁ = $\frac{I}{m x^2 + I} \ w_o$

Let's repeat for the platform with the bananas and the monkey is the one that falls for x₂ = 1.73 m

initial instant. The platform and bananas alone

L₀ = I₁ w₁

I₁ = (m x² + I)

final instant. After the crash

L_f = I w

L_f = (I₁ + M x₂²) w

the moment is preserved

L₀ = L_f

(m x² + I) w₁ = ((m x² + I) + M x₂²) w

(m x² + I) w₁ = (I + m x² + M x₂²) w

we substitute

w = $\frac{m x^2 +I}{I + m x^2 + M x_2^2} \ \frac{I}{m x^2 + I} \ w_o$

w = $\frac{I}{I + m x^2 + M x_2^2} \ w_o$

the moment of inertia of a circular disk is

I = ½ m_p x₂²

we substitute

w = $\frac{ \frac{1}{2} m_p x_2^2 }{ \frac{1}{2} m_p x_2^2 + M x_2^2 + m x^2} \ \ w_o$

let's calculate

w = $\frac{ \frac{1}{2} \ 97.3 \ 1.73^2 }{ \frac{1}{2} \ 97.3 \ 1.73^2 + 21.9 \ 1.73^2 + 9.67 \ 0.45^2 } \ \ 1.67$

w = $\frac{145.60 }{145.60 \ + 65.54 \ + 1.958} \ \ 1.67$

w = 1.14 rad / s

5 0

3 years ago

A body of density 9.0cm appears to have mass 27.0g in a liquid of density 1.2gcm. What is the volume of the solid?

Alika [10]

Answer:

v * 9.0 = 27.0 + (v * 1.2)

7 0

3 years ago

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find

nikitadnepr [17]

Answer:

a) x = v₀ₓ t , y = $v_{oy}$ t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis

-W = m $a_{y}$

-m g = m $a_{y}$

$a_{y}$ = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

x = v₀ₓ t + ½ a t²

As the acceleration on this axis is zero

x = v₀ₓ t

Y Axis

y = $v_{oy}$ t + ½ $a_{y}$ t²

The acceleration on this axis is –g

y = $v_{oy}$ t - ½ g t²

B) to find the maximum value of distance r

r =√ x² + y²

r = √( v₀ₓ² t² + ( $v_{oy}$ t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

dr / dt = 0

0 = ½ 1/√( v₀ₓ² t² + ( $v_{oy}$ t + ½ g t²)² (v₀ₓ² 2t + 2 ( $v_{oy}$ t - ½ g t²)( $v_{oy}$ - ½ g2t)

We simplify this expression

0 = v₀ₓ² 2t + 2 ( $v_{oy}$ t - ½ g t²) ( $v_{oy}$ - ½ g2t)

-v₀ₓ² t =( $v_{oy}$ t - ½ g t²) ( $v_{oy}$ - ½ g2t)

-v₀ₓ² t = $v_{oy}$ ² t -3/2 gt² $v_{oy}$ + 1/2 g² t³

½ g² t²- 3/2 g $v_{oy}$ t = $v_{oy}$ ² + v₀ₓ²

Let's use trigonometry to find go and vox

sin θ = $v_{oy}$ / v₀

cos θ = vox / v₀

$v_{oy}$ = v₀ sin θ

v₀ₓ = v₀ cos θ

We replace

½ g² t² -3/2 g $v_{oy}$ t = v₀ (sin² θ + cos² θ)

g t² - 3v₀ sin θ t = 2 v₀/g

The time is maximum for the angle is zero

8 0

3 years ago

1 Item 1 Item 1 1.25 points circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. T

Elena L [17]

Answer:

174.85 W

Explanation:

Area of plate = 3.14 x (15x 10⁻²)²

= 706.5 x 10⁻⁴ m²

heat being radiated by convection = 12 x 706.5 x 10⁻⁴ ( 180 - 15 )

= 139.88 W. This energy needs to be fed by heat source to maintain a constant temperature of 180 degree.

If power of electric source is P

P x .8 = 139.88

P = 139.88 / .8

= 174.85 W

6 0

3 years ago