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stealth61 [152]
3 years ago
7

PLEASE HELP URGENT!!!!!!!!!!!!!

Physics
1 answer:
Nimfa-mama [501]3 years ago
4 0

c i think sorry if wrong

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A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of
Fudgin [204]

Answer:

Part a)

v_f = v_x = 32.77 m/s

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

v_x = 40 cos35 = 32.77 m/s

v_y = 40 sin35 = 22.94 m/s

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

v_f = 32.77 m/s

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

\Delta y = 0

0 = v_y t - \frac{1}{2}gt^2

T = \frac{2v_y}{g}

T = \frac{2(22.94)}{9.81} = 4.68 s

7 0
3 years ago
What is the volume of an object that has Mass=100g and Density= 745g/mL.
lana66690 [7]

Answer:0.13ml

Explanation:

6 0
3 years ago
A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak
Zina [86]

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

I'' = I' Cos^{2}\theta

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 12.5\times Cos^{2}15

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0
3 years ago
A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri
Drupady [299]

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

KE = I/2*\omega_o^2

The inertia is:

I=m/2*R^2=0.072kg.m^2

So, the kinetic energy will be:

KE = 888.26J

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf    where Kf=0;   Ko = 888.26J;  and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R   where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

-888.26=-0.96*π*N

N=294.5 turns

6 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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