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uranmaximum [27]

3 years ago

7

Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curv

e O B. A car slowing down as it goes around a curve 23 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road

1 answer:

ki77a [65]3 years ago

7 0

Answer:

Pluto revolves around sun with constant speed in a circular orbit, so it is an example of uniform circular motion.

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The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of

Karolina [17]

Answer:

ar = 5.86*10^-3 m/s^2

Explanation:

In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.

You use the following formula:

$v=\sqrt{\frac{GM_s}{r}}$ (1)

G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2

Ms: Sun's mass = 1.98*10^30 kg

r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m

Furthermore, you take into account that the radial acceleration is given by:

$a_r=\frac{v^2}{r}$ (2)

You replace the equation (1) into the equation (2) and replace the values of all parameters:

$a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}$

The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2

7 0

3 years ago

Which type of map would you most likely use to locate mineral deposits? A. topographic B. geologic C. satellite D. hazard

qwelly [4]

I think you would be using a topographic Map, So the answer should be A

3 0

3 years ago

The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around

zubka84 [21]

Answer:

8 electrons in the third energy level

Explanation:

From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0

4 years ago

An airplane is flying through the air at a speed of 150 mph at a heading of 60 degrees. if the wind is blowing at 20 mph from th

Crazy boy [7]

Answer:

Explanation:

Given

speed of plane 150 mph

heading $60^{\circ}$

wind is blowing at 20 mph from south-east

velocity of plane w.r.t. wind

$v_{pw}=150(\cos 60\hat{i}+\sin 60\hat{j})$

$v_{w}=20(\cos 45\hat{i}-\sin 45\hat{j})$

$v_{p}=v_{pw}+v_{w}$

$v_{p}=(150\cos 60+20\cos 45)\hat{i}+(150\sin 60-20\sin 45 )\hat{j}$

$v_{p}=89.14\hat{i}+115.76\hat{j}$

$|v_{p}|=146.103 m/s$

(b)after 30 minutes

Plane traveled in x direction $=89.14\times 0.5=44.57 miles$

Plane in Y direction $=115.76\times 0.5=57.88 miles$

Total distance $=\sqrt{x^2+y^2}=73.05 miles$

direction $tan\theta =\frac{115.76}{89.14}=1.298$

$\theta =52.38^{\circ}$ w.r.t east

4 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

3 years ago