Ask question

Ask question

All categories

3 months ago

12

An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight

downward. What is the component of the apple's impact velocity parallel to the surface of the slope?
a.5.5 m/s
b.8.7 m/s
c.12 m/s
d.15 m/s

1 answer:

serg [3.5K]3 months ago

7 0

An apple strikes the ground at a velocity of 16.2 m/s.

The angle between the velocity of the apple and a line normal to the inclined surface is 20 degrees.

The parallel and perpendicular components of its velocity concerning the surface are as follows:

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

This gives us:

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

The velocity along the inclined plane measures 5.5 m/s.

You might be interested in

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

serg [3582]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Find out more about average velocity:

6 0

3 months ago

A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [3103]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

4 0

2 months ago

The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet

Keith_Richards [3271]

The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.

7 0

2 months ago

Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w

Keith_Richards [3271]

Solution:

$Em_{f}$ / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

K = ½ m v²

U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

Em₀ = K = ½ m v²

At its highest point

$Em_{f}$ = U = mg and

Now let's compute

Em₀ = ½ m 3.6²

Em₀ = m 6.48

$Em_{f}$ = m 9.8 × 0.2

$Em_{f}$ = m 1.96

Thus the energy lost is given by:

$Em_{f}$ / Em₀ = m 1.96 / m 6.48

$Em_{f}$ / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

8 0

2 months ago