PHYSICS HIGH SCHOOL

A worker kicks a flat rock lying on a roof. The rock slides up the incline 10.0 m to the apex of the roof, and flies off the roof as a projectile. What maximum height (in m) does the rock attain? Assume air resistance is negligible, vi = 15.0 m/s, μk = 0.425, and that the roof makes an angle of θ = 42.0° with the horizontal. (Assume the worker is standing at y = 0 when the rock is kicked.)

Answers

Answer 1

Answer:

The maximum height that the rock attains is : 7.42 m

Given data :

Vi = 15.0 m/s

uk = 0.425

angle made by roof ( θ ) = 42.0°

Calculating the maximum height reached by the rock

First step : calculate the deceleration of the rock

a = g*sinθ + uk * g * cosθ

= 9.8 * sin 42° + 0.425 * 9.8 * cos 42°

= 9.65 m/s²

Applying motion formula

v² = u² + 2as

= 15² - 2 * 9.65 * 10

v = 5.66 m/s

Initial height attained ( h₁ ) = 10 sin θ

= 10 * sin 45°

= 6.69 m

considering vertical height

V (y) = v sin θ

= 5.66 sin 42

= 3.79 m/s

Next step : Calculate height ( h₂ )

v² = u² + 2 a s

0 = 3.79² - 2 * 9.8 * h₂

therefore : h₂ = 0.73 m

The maximum height attained by the rock ( H ) = h₁ + h₂

= 6.69 + 0.73 = 7.42 m

Hence we can conclude thatThe maximum height that the rock attains is : 7.42 m

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Answer 2

Answer:

Answer:

h = 7.42 m

Explanation:

deceleration of the rock

$a = g sin \theta + \mu_kgcos \theta$

$a = 9.8 sin 42^0+0.425* 9.8 * cos 42^0$

a = 9.65 m/s²

using formula

v² = u² + 2 a s

v² = 15² - 2×9.65 × 10

v = 5.66 m/s

the height attained is

h₁ = 10 sin θ

= 10 sin 42

= 6.69 m

now with vertical velocity it will reach to the height h₂

v y = v sin θ

= 5.66 sin 42

= 3.79 m/s

height is

v² = u² + 2 a s

0 = 3.79² - 2 × 9.8 ×h₂

h₂ = 0.73 m

the maximum height is

h = h₁ + h₂

= 6.69 + 0.73

h = 7.42 m

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Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on the other to be LaTeX: 1.35\times10^{-4}N1.35 × 10 − 4 N when they are 20 cm apart. Accidentally, one the the experimenters causes the balls to collide and then repositions them 20 cm apart . Now the repulsive force is found to be LaTeX: 1.406\times10^{-4}N1.406 × 10 − 4 N. What are the initial charges on the two metal balls?

Answers

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=(1)/(4\pi\spsilon_0)(q_1q_2)/(d^2)-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$(1)/(4\pi\spsilon_0)=9*10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35*10^(-4)$ N

So, from equation (i)

$1.35*10^(-4)=9*10^9(q_1q_2)/((0.2)^2)$

$\Rightarrow q_1q_2=6*10^(-16)\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$(q_1+q_2)/(2)$

d=20cm=0.2m, and $F_c=1.406*10^(-4)$ N

So, from equation (i)

$1.406*10^(-4)=9*10^9(\left((q_1+q_2)/(2)\right)^2)/((0.2)^2)$

$\Rightarrow (q_1+q_2)^2=2.50*10^(-15)$

$\Rightarrow q_1+q_2=\pm5* 10^(-8)$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5* 10^(-8)\;\cdots(iii)$

$\Rightarrow q_1=5* 10^(-8)-q_2$

The equation (ii) become:

$(5* 10^(-8)-q_2)q_2=6*10^(-16)$

$\Rightarrow -(q_2)^2+5* 10^(-8)q_2-6*10^(-16)=0$

$\Rightarrow q_2=3*10^(-8), 2*10^(-8)$

From equation (iii)

$q_1=2*10^(-8), 3*10^(-8)$

So, the magnitude of initial charges on both the sphere are $3*10^(-8)$ Coulombs $=0.03 \mu C$ and $2*10^(-8)$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

What is the rate at which electrical energy is converted to another form of energy called ?

Answers

the rate is called power and its unit is voltage

Final answer:

The rate at which electrical energy is converted to another form of energy is called power.

Explanation:

In an electric circuit, electrical energy is continuously converted into other forms of energy. For example, when a current flows in a conductor, electrical energy is converted into thermal energy within the conductor. This process is called power. Power is the rate at which energy is transferred or work is done, and is measured in watts.

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An electric fence displays a warning sign about voltage and amperage of the fence. How would amperage and voltage affect the power of the fence?

Answers

Yes it is true that electrical fences displays a warning about voltage and amperage of the fence. It has to be first understood that power is directly proportional to voltage as well as current. So higher the voltage or higher the current flowing through the fenses, higher will be the power. So any person touching the fence should always be aware of the voltage and current flowing through the fence, otherwise it can be very deadly. Sometimes it can be seen that a high voltage is being passed through the fence and the current passed is comparitively low. This will not decrease the total power flowing through the fence.

Which element was discovered by joseph priestley to be what his contemporaries called ""dephlogisticated air""?

Answers

Answer:

oxygen

During his lifetime, Priestley's considerable scientific reputation rested on his invention of soda water, his writings on electricity, and his discovery of several "airs" (gases), the most famous being what Priestley dubbed "dephlogisticated air" (oxygen).

Explanation:

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