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MrRissso [65]
2 years ago
14

A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec

tile at the moment it strikes the wall?
A. 29.4 m/s down
B. 29.4 m/s east
C. 29.4 m/s at 45(degrees) down from horizontal toward the east
D. 41.6 m/s at 45(degrees) down from horizontal toward the east
Physics
1 answer:
Drupady [299]2 years ago
6 0

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s  (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

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The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t&gt;0. Exp
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2 years ago
Read 2 more answers
a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio
Katena32 [7]

Answer:

a) factor b=\sqrt{2}

b) factor b=\frac{1}{2}

c) factor b=1

d) factor b=1

Explanation:

Time period of oscillating spring-mass system is given as:

T=\frac{1}{f}

T={2\pi} \sqrt{\frac{m}{k} }

where:

f= frequency of oscillation

m= mass of the object attached to the spring

k= stiffness constant of the spring

a) <u>On doubling the mass:</u>

  • New mass, m'=2m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k} }

T'=2\pi\sqrt{\frac{2m}{k} }

T'=\sqrt{2}\times  2\pi\sqrt{\frac{m}{k} } }

T'=\sqrt{2} \times T

where the factor b=\sqrt{2} as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, k'=4k

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T

where the factor  b=\frac{1}{2}  as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, k'=4k

New mas, m'=4m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T

where factor b=1 as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor b=1

7 0
1 year ago
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