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Licemer1 [7]
2 years ago
7

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

t other launch angle will the projectile have the same range?
Physics
1 answer:
VLD [36.1K]2 years ago
3 0

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
2 years ago
How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​
Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0
1 year ago
A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.
Sholpan [36]

Answer:

5cm east& 1cm west from A

Explanation:

https://brainly.ph/question/2753392

7 0
1 year ago
Two objects interact with each other and with no other objects. Initially object A has a speed of 5 m/s and object B has a speed
Radda [10]

Answer:

We can conclude that there is a decrease in kinetic energy of the particles due to their elastic collision, since kinetic energy is directly proportional to squared velocity of the particles.

Explanation:

Given:

initial velocity of particle A, Ua = 5m/s

initial velocity of particle B, Ub = 10 m/s

final velocity of particle A, Va = 4m/s

final velocity of particle B, Vb = 7m/s

For particle A:

The final velocity is 1 less than the initial velocity.

For particle B:

The final velocity is 3 less than the initial velocity.

We can conclude that there is a loss in kinetic energy due to elastic collision of the two particles, since kinetic energy is directly proportional to squared velocity of the particles. A decrease in velocity means decrease in kinetic energy.

4 0
2 years ago
Identify the method of thermal energy transfer at work in hot air balloons. Explain how thermal energy is transferred in this sc
yan [13]
Thermal energy in the form of fire is generated by the combustion of fuel. Due to the tendency of hot air to rise upward, the heat generated rises to fill the space of the balloon. One this space is full of trapped hot air, the heat's tendency to rise causes the hot air balloon to be lifted into the air. 
8 0
2 years ago
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