Ask question

Ask question

All categories

2 years ago

10

A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1

40 m above the launch point. (a) Calculate the horizontal component of its velocity. (b) Calculate the vertical component of its velocity just after launch. (c) At one instant during its flight the vertical component of its velocity is found to be 65 m/s. At that time, how far is it above or below the launch point

1 answer:

SCORPION-xisa [38]2 years ago

5 0

Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

You might be interested in

Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by th

Andrej [43]

Answer:

The charge carried by the droplet is $1.330\times10^{-19}\ C$

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude $5.92\times10^4\ N/C$ points straight down and if the mass of the droplet is $2.93\times10^{-15} kg$

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2$

$a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}$

$a=2.688\ m/s^2$

We need to calculate the charge carried by the droplet

Using formula of electric filed

$E=\dfrac{F}{q}$

$q=\dfrac{ma}{E}$

Put the value into the formula

$q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}$

$q=1.330\times10^{-19}\ C$

Hence, The charge carried by the droplet is $1.330\times10^{-19}\ C$

8 0

2 years ago

choli [55]

Answer:

B. Acceleration in the direction of motion speeds you up

Explanation:

Acceleration is defined as when something gains speed. For example, when a car speeds up. Deceleration is when the car slows down, and looses speed. When defining these terms, think of a car going faster, then slowing down at a red light.

3 0

2 years ago

A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

Andreas93 [3]

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

$p=m_pv_p+m_bv_b=0$

Therefore,

$v_p=-\frac{m_b}{m_p} v_p$

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

$v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s$

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

8 0

2 years ago

You are on the south bank of the river in your canoe you need to reach the north bank you know that you can row your canoe at 2

Anna71 [15]

Answer:

(θ) = 60°

Explanation:

Given:

Speed of canoe Vc = 2 m/s

Speed of River Vr = 1 m/s

Computation:

Vc (Cosθ) = Vr

2 (Cosθ) = 1

(Cosθ) = 1 / 2

(Cosθ) = (Cos60)

(θ) = 60°

3 0

2 years ago

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

$\theta=274rev$

Explanation:

From the question we are told that:

Angular velocity $\omega=510rpm$

Mass $m=40.kg$

Diameter d $75=>0.75m$

Off Time $t=40.0s$

Oscillation at Power off $N=210$

Generally the equation for Angular displacement is mathematically given by

$\theta_{\infty}=\frac{w+w_0}{t}t$

$w=\frac{2*\theta_{\infty}}{t}-w_0$

$w=\frac{28210}{40*(\frac{1}{60})}-510$

$w=120rpm$

Generally the equation for Time to come to rest is mathematically given by

$t=(\frac{\omega_0}{\omega_0-\omega})t$

$t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})$

$t=0.87min$

Therefore Angular displacement is

$\theta =(\frac{120+510}{2})0.87$

$\theta=274rev$

6 0

2 years ago