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As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

vlada-n [284]

Answer:

$a=-0.33\ m/s^2$

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

$\displaystyle a=\frac{\Delta v}{t}$

Or, equivalently

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

$\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2$

The negative sing of a indicates there is deceleration or decreasing speed

7 0

3 years ago

Pls ans 10 no. From laws of motion

Vlad1618 [11]

Answer:

40N

Explanation:

Since both weights are connected to one string, you can say that the tensions above each are equal to each other.

If you do the sum of forces for the 4kg mass, then the tension comes out to 40N (if we take gravity to be 10m/s²). But that seemed too good to be true, so I decided to do the work for the 7kg mass as well [which included finding the normal force (N) and plugging it into the sum of forces for the 7kg mass] to find that it also gives 40N as the answer.

If I were to put my process into steps:

Write out the sum of Forces for both masses
Set them equal to each other to find normal force (because this is the only unknown)
Calculate and compare the two tensions to see if they are equal

*This all seems to line up perfectly, but do let me know if my answer doesn't match up with what you might find to he the answer later on.

4 0

3 years ago

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

7 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

3 years ago

Easy one here pls help thanks so much

Kisachek [45]

The answer is B. Nutrients.

8 0

2 years ago

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This is a voice recording of the text of a book that you listen rather than read.​

This is a voice recording of the text of a book that you listen rather than read.