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Anna35 [415]
2 years ago
7

A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat

e the maximum height the ball reaches. (Use 10 m/s2 as the acceleration of gravity.) 101 Incorrect: Your answer is incorrect.
Physics
1 answer:
mihalych1998 [28]2 years ago
6 0

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

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You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t
shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0
3 years ago
What is the momentum of a 750 kg car traveling at a velocity of 25 m/s north?
olga_2 [115]

Answer:

18750 kg-m/s

Explanation:

Momentum = mass x velocity 

3 0
2 years ago
Read 2 more answers
A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
balandron [24]

Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

Hence:

\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

In other words, the angle of refraction in the glass would be approximately 28^{\circ}.

7 0
2 years ago
Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2
crimeas [40]

Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

         ∑ τ = I α

         

we will assume that the counterclockwise turns are positive

         F₁  0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

        I = ½ M R₂²

         

        α = (F₂ R₂ - F₃ R₃)  \frac{2}{M R_2^2}

let's calculate

        α = (24  0.22 - 13  0.10) \frac{2}{12 \ 0.22^2}2/12 0.22²

        α = 13.7 rad / s²

6 0
3 years ago
A 63 kg kg person starts traveling from rest down a waterslide 6.0 mm above the ground. At the bottom of the waterslide, it then
gladu [14]

Answer:

change waterslide according to question. and you are good to go. check photo for solve

5 0
3 years ago
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