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ycow [4]
3 years ago
10

How to find initial velocity in projectile motion problems when you are not given the vlocity?

Physics
1 answer:
Papessa [141]3 years ago
7 0
Vi= square root vi^2 -2ad
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What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a
skad [1K]
The answer would be a=5 m/s^2
I hope this helps you, have a great day!
8 0
3 years ago
Read 2 more answers
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?
Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

5 0
2 years ago
Read 2 more answers
I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t
dezoksy [38]
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where L is the length of the string, T the tension and m its mass. If  we plug the data of the problem into the equation, we find
f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz

The wavelength of the standing wave is instead twice the length of the string:
\lambda=2 L= 2 \cdot 24 m=48 m

So the speed of the wave is
v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s
7 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
3 years ago
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