I NEED HELP ASAP!!!

A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?

nata0808 [166]

Answer:

v=u+at

v=21+1.4(11)

v=21+15.4

v=36.4

3 0

3 years ago

A string under a tension of 68 N is used to whirl a rock in a horizontal circle of radius 3.7 m at a speed of 16.53 m/s. The str

alekssr [168]

Answer:

$F = 5253.7 N$

Explanation:

As we know that tension force in the string will be equal to the centripetal force on the string

so we will have

$T = \frac{mv^2}{L}$

now we have

$68 = \frac{m(16.53^2)}{3.7}$

now we have

$68 = 73.8 m$

$m = 0.92 kg$

now when string length is 0.896 m and its speed is 71.5 m/s then we will have

$F = \frac{mv^2}{r}$

$F = \frac{0.92(71.5^2)}{0.896}$

$F = 5253.7 N$

8 0

3 years ago

Read 2 more answers

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

$tangent( O ) = \frac{opposite}{adjacent}$

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

$tangent( 4) = \frac{53 feets}{DB}$

$DB= \frac{53 feets}{tangent( 4)}$

$DB= 757,93 feets$

This its equal to the distance from the boat to the shoreline.

4 0

3 years ago

What are the wavelength ranges for the following? (a) the AM radio band (540–1600 kHz) maximum wavelength m minimum wavelength m

Pie

Answer:

Explanation:

a ) AM radio band (540–1600 kHz)

frequency = 540 kHz = 540 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 540 x 10³

= 555.55 m

frequency = 1600 kHz = 1600 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 1600 x 10³

= 187.5 m

maximum wavelength = 555.55 m

minimum wavelength = 187.5 m

b )

AM radio band (88 - 108 MHz)

frequency = 88 MHz = 88 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 88 x 10⁶

= 3.41 m

frequency = 108 MHz = 108 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 108 x 10⁶

= 2.78 m

maximum wavelength = 3.41 m

minimum wavelength = 2.78 m

3 0

3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago