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Fynjy0 [20]
2 years ago
14

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, is this an elastic collision?
A) Yes, since momentum is conserved.
B) No, since momentum is not conserved.
C) Yes, since kinetic energy is conserved.
D) No, since kinetic energy is not conserved.
E) Not enough information.
Physics
1 answer:
TEA [102]2 years ago
5 0

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

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In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to
dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, a=\dfrac{v-u}{t} ....(1)

The third equation of kinematics is as follows :

v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s

Hence, it will take 25 seconds to increase the speed.

6 0
2 years ago
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
An unspecified force causes a 0.20-kg object to accelerate at 0.40 m/s2. If 0.30 kg is added to the 0.20-kg object and the force
Naddik [55]

Answer:

a = 0.16

Explanation:

given,

mass of the object 1 = 0.2 kg

mass of the object 2 = 0.3 kg

acceleration when force is on 0.2 kg = 0.4 m/s²

acceleration when both mass are combine = ?

F = m a

F = 0.2 × 0.4

F = 0.08 N

force acting is same and total mass  = 0.2 + 0.3 = 0.5 Kg

F = m a

a = \dfrac{F}{m}

a = \dfrac{0.08}{0.5}

a = 0.16 m/s²

the acceleration  acting when both the body is attached is a = 0.16

4 0
2 years ago
A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0
2 years ago
Read 2 more answers
Dee is on a swing in the playground. the chains are 2.5 m long, and the tension in each chain is 450 n when dee is 55 cm above t
Leno4ka [110]
Refer to the diagram shown below.

From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
cos θ = 1.95/2.5 = 0.78
θ = cos⁻¹ 0.78 = 38.74°

From the free body diagram, the tension in the chain is 450 N.
F is the centripetal force,
W is Dee's weight.

The components of the tension are
Horizontal component = 450 sin(38.74°) = 281.6 N, acting left.
Vertical component = 450 cos(38.74°) = 351.0 N, acting upward.

Answers:
Horizontal: 281.6, acting left.
Vertical: 351.0 N, acting upward.

8 0
2 years ago
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