100 POINTS WILL GIVE BRAINIEST TO BEST ANSWER!!!!!!!!!!!!!<br><br> see below

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

A center-seeking force related to acceleration is _______ force.

Gennadij [26K]

<span>A center-seeking force related to acceleration is centripetal force. The answer is letter A. The rest of the choices do not answer the question above.</span>

4 0

3 years ago

A small car with mass m and speed 2v and a large car with mass 2m and speed v both travel the same circular section of an unbank

Alenkasestr [34]

Answer:

(D) F/2

Explanation:

Since the circular section is unbanked, the centripedal acceleration acting on each of the cars is

$a_s = \frac{v_s^2}{r} = \frac{(2v)^2}{r} = \frac{4v^2}{r}$

$a_l = \frac{v_l^2}{r} = \frac{v^2}{r}$

Therefore the centripetal force on each car

$F_s = m_sa_s = \frac{4mv^2}{r}$

$F_l = m_la_l = \frac{2mv^2}{r}$

Since $F_s = 2F_l$ this means the friction force required to keep the large car on the road is only half of the friction force required to keep the small car on road

So (D) F/2 is the correct answer

5 0

3 years ago

g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b

Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c) Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

Em₀ = K = ½ m v²

final point. Higher

Em_f = U = mg h

Em₀ = Em_f

Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

Em₁ = K = ½ m v²

final point. Higher

Em_f = U = mg h

Em₁ = Em_f

Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0

2 years ago

(a) Without the wheels, a bicycle frame has a mass of 6.75 kg. Each of the wheels can be roughly modeled as a uniform solid disk

adelina 88 [10]

Answer:

a) Ktotal = 71.85 J

b) Ktotal = 71.85 J

Explanation:

a) The total kinetic energy is that of the total mass of the bicycle plus the rotational kinetic energy of the two wheels. The linear speed of the circumference of the wheels matches the forward speed of the bicycle, so their angular speed is

ω = v/r

The moment of inertia of one solid disk bicycle wheel is

I = 0.5*m₂*r²

And the rotational kinetic energy of one wheel is

Kr = 0.5*I*ω² = 0.5*(0.5*m₂*r²)*(v/r)² = 0.25*m₂*v²

The total kinetic energy is then that of the frame and wheels plus the rotational kinetic energy.

Ktotal = 0.5*(m₁ + 2*m₂)*v² + 2*(0.25*m₂*v²)

⇒ Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 6.75 Kg

m₂ = 0.820 kg

v = 3.95 m/s

then

⇒ Ktotal = 0.5*(3.95 m/s)²*(6.75 Kg + 3*0.820 kg)

⇒ Ktotal = 71.85 J

b) We can apply the same equation obtained before

⇒ Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 675 Kg

m₂ = 82.0 kg

v = 0.395 m/s

then

⇒ Ktotal = 0.5*(0.395 m/s)²*(675 Kg + 3*82 kg)

⇒ Ktotal = 71.85 J

3 0

3 years ago

100 POINTS WILL GIVE BRAINIEST TO BEST ANSWER!!!!!!!!!!!!! see below