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4 years ago

An object has a kinetic energy of 14 J and a mass of 17 kg , how fast is the object moving?

2 answers:

lisabon 2012 [21]4 years ago

6 0

$v ^{2}$ = Joules ÷ (0.5×Kilograms)

14J ÷ 8.5 = 1.64705882

Remember, 1.64705882 = v², so we need to find the square root.

The square root of 1.64705882 is 1.283377894464448

Hope this helps!

valkas [14]4 years ago

4 0

Answer:

1.283377894464448

Explanation:

Big brain

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A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a

Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

$F_N = F_C - F_G = m\frac{v^{2} }{r} - mg$

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

$E_{tot}=mgh + \frac{1}{2} mv^{2}$

At the top of the hill:

$E_{tot}= mgh_{hill}$

At the top of the loop:

$E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}$

Setting both energies equal and canceling the mass m gives:

$gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}$

Solving for v:

$v^{2} = 2g(h_{hill}-h_{loo}_p)$

Using v in the first equation:

$F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg$

$F_N = 955.5N$

7 0

3 years ago

Can someone help me pls

Dmitry_Shevchenko [17]

First question: 800J
Second question: 20.4m

4 0

3 years ago

Which of the following is not an advantage that attractive people enjoy in interpersonal relationships?

Licemer1 [7]

Psychology on Egenuity Oct 5th 2018 says answer is C

5 0

4 years ago

Read 2 more answers

Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to

katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

$OB=0.40 km$

And the Max distance towards bookstore is,

$OA=3.65 km$

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

$AB=\sqrt{OB^{2}+OA^{2}}$

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

$AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km$

Therefore the distance between there destination is 3.67 km.

6 0

3 years ago