A heavy object is dropped from a vertical height of 8.0 m. What is its speed when it hits the ground?

A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North

ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

$magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km$

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km

3 0

3 years ago

Can anyone help me with these questions? TIA!<br> (Don’t actually answer please! :) )

nataly862011 [7]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

$\huge\text{Graphs:}$

The graph of

• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1

• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2

<h2>_____________________________________ </h2><h2>Question 8: </h2>

$\Large\textbf{Diode:}$

A diode is a device that allows current to flow in only one direction.

$\Large\textbf{Forward and Reverse Biasing:}$

Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)

Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.

$\Large\textbf{Answer to the Question "Resistance"}$

The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.

<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>

8 0

3 years ago

A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di

slava [35]

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ = 14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ = $\frac{n_1sin\theta_1}{n_2}$ = $\frac{1.003*0.251}{1.33}$ = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = $\frac{4}{x_2}$ or x₂ = $\frac{4}{tan\theta_2 }$ = $\frac{4}{tan(10.86)}$ = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

5 0

3 years ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

ValentinkaMS [17]

The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

<h3>Acceleration of the box</h3>

The acceleration of the box is calculated as follows;

vf² = vi² + 2as

a = (vf² - vi²)/2s

a = (11.5² - 13²) / (2 x 8.5)

a = -2.16 m/s²

<h3>Time of motion of the box</h3>

The time taken for the box to travel is calculated as follows;

a = (vf - vi)/t

t = (vf - vi) / a

t = (11.5 - 13) / (-2.16)

t = 0.69 s

<h3>Average power supplied by the friction</h3>

P = Fv

P = (ma)(vf - vi)

P = (1 x -2.16) x (11.5 - 13)

P = 3.24 W

Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

Learn more about average power here: brainly.com/question/19415290

#SPJ1

7 0

2 years ago

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

5 0

3 years ago