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lakkis [162]
3 years ago
8

Meg walks with a velocity of 0.9 m/s west. She does so while riding on a train that is traveling with a velocity of 2.7 m/s east

. What is Megs velocity? A. 3.6 m/s west B. 1.8 m/s west C. 1.8 m/s east D. 3.6 m/s east
Physics
1 answer:
Shkiper50 [21]3 years ago
6 0
<span>Velocities are vectors so we can add them!

Let's let +x be East and -x be West.

 -0.9 + 2.7 = 1.8
 
Since our answer is positive that means East so the answer is C.</span>
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Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
Charra [1.4K]

Answer:

Time of flight  A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight  A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight  A is greatest .

8 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
What is the definition of input force?
andrezito [222]

Answer:

The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.

Explanation:

5 0
2 years ago
Read 2 more answers
in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo
guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

4 0
3 years ago
Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball
Likurg_2 [28]

Answer:

k = 4422.35  KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k        

k=Spring constant

So we can say that

( 2 π f)² =  m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³  k

33167.69 = 7.5 x 10⁻³  k

k=4422.35 x  10³ N/m

k = 4422.35  KN/m

Therefore spring constant will be 4422.35  KN/m

4 0
3 years ago
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