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torisob [31]
3 years ago
13

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

grees? A. 15 degrees. B. 65 degrees. C. 70 degrees. D. 80 degrees. E. 90 degrees.
Physics
1 answer:
snow_tiger [21]3 years ago
5 0

As we know that range of the projectile motion is given by

R = \frac{v^2 sin(2\theta)}{g}

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

\theta, 90 - \theta

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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The contraction of muscles in the stomach walls
soldi70 [24.7K]
<span>The contraction of muscles in the stomach walls is called peristalsis. 
Peristalsis works to physically break down food and move it forward.

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8 0
3 years ago
When a metal wire has an electric field applied, electric current flows. a) If we consider the path the electrons take through t
Zielflug [23.3K]

Answer:

Check the explanation

Explanation:

(a) When the electric field is applied to the ends of the wire, electrons move through the wire because the resistance of the wire is less than the resistance of the surrounding air. Resistance is the obstruction to the flow of electrons hence electrons flow through the path having minimum resistance.

(b) As the thermal energy increases, Kinetic energy of electrons as well as that of positive ions also increase. Hence the mean time between collisions of electrons with ions is reduced due to movement of ions also. That is why Resistance of the wire increases.

8 0
3 years ago
4 points
zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

I = k\frac{1}{r^2}

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

I_2 = k\frac{1}{(2r)^2}

Dividing both the equation we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4}  = 25lx

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

I_2 = k\frac{1}{(3r)^2}

Dividing equation 1 and 3 we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9}  = 11.11lx

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0
3 years ago
If you were trying to kill a fish witha a spear, woukd you ain where you see the fish?
patriot [66]
No, you can't aim where you see the fish, because refraction will change the position of the fish and make it appear in a different position from where it actually is.

Hope this helps!
8 0
3 years ago
Read 2 more answers
A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.
miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0
1 year ago
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