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Licemer1 [7]
2 years ago
7

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

t other launch angle will the projectile have the same range?
Physics
1 answer:
VLD [36.1K]2 years ago
3 0

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

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Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

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Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

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im guessing it's 5 m/s

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The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
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</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
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<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
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3 0
2 years ago
Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for th
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6 0
2 years ago
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A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

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from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
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