All categories

3 years ago

A solar-powered car has a kinetic energy of 110250 J. Its mass is 180 kg. Work out how fast the car is travelling.

Physics

2 answers:

vesna_86 [32]3 years ago

4 0

Explanation:

Kinetic Energy 1 upon 2 into mass into velocity square

1 upon 2 into 180 into velocity square=110250

v^2=110250 multiple by 2 and divided by 180

And take square root

shtirl [24]3 years ago

3 0

Answer:

35 m/s

hope this helpedd

You might be interested in

Help needed pls and thxs

algol13

Erosion:
A.) Is the same as weathering

3 0

4 years ago

The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined pl

Scorpion4ik [409]

Given :-

A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
$\mu_s = 0.3 \ \& \ \mu_k = 0.2$

To Find :-

Would the block move ?
If it moves what is its speed after it has descended a distance of 5m down the plane .

Solution :-

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,

$\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)$

Substitute ,

$\longrightarrow \theta_{repose}= tan^{-1}( 0.2)$

Solve ,

$\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}$

Hence ,

$\longrightarrow\theta_{plane}>\theta_{repose}$

Hence the block will slide down .

Now assuming that block is released from the reset , it's initial velocity will be 0m/s .

And the net force will be ,

$\longrightarrow F_n = mgsin53^o - \mu_k N$

Substitute, N = mgcos53⁰ ( see attachment)

$\longrightarrow ma_n = mgsin53^o - \mu_k mgcos53^o$

Take m as common,

$\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)$

Simplify ,

$\longrightarrow a_n = gsin53^o - \mu_k g cos53^o$

Substitute the values of sin , cos and g ,

$\longrightarrow a_n = 10( 0.79 - 0.2 (0.6))$

Simplify ,

$\longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}$

Now using the Third equation of motion namely,

$\longrightarrow2as = v^2-u^2$

Substituting the respective values,

$\longrightarrow2(6.7)(5) = v^2-(0)^2$

Simplify and solve for v ,

$\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}$

Hence the velocity after covering 5m is 8.18 m/s .

7 0

3 years ago

What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit

Tresset [83]

The highest frequency sound to which the machine can be adjusted is :

4179.33 Hz

Given data :

Pressure = 10 Pa

Speed of sound = 344 m/s

Displacement altitude = 10⁻⁶ m

<h3>Determine the highest frequency sound ( f ) </h3>

applying the formula below

Pmax = $B(\frac{2\pi f}{v}) A$ --- ( 1 )

Therefore :

f = ( Pmax * V ) / $2\pi \beta A$

= ( 10 * 344 ) / 2 $\pi$ * 1.31 * 10⁵ * 10⁻⁶

= 4179.33 Hz

Hence we can conclude that The highest frequency sound to which the machine can be adjusted is : 4179.33 Hz .

Learn more about Frequency : brainly.com/question/25650657

Attached below is the missing part of the question

A loud factory machine produces sound having a displacement amplitude in air of 1.00 μm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.31×105 Pa. The speed of sound in air is 344 m/s. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?

4 0

2 years ago

When Jacob bats a baseball with a net force of 4.719 N, it accelerates 33 m/s2. What is the mass of the baseball?

andreev551 [17]

Answer:

Mass=0.143 kg

Explanation:

From the Newton's second law of motion the force applied to an object is directly proportional to the acceleration produced and the acceleration is in the direction of the force.

F=ma where F is the force, m is the mass of the object and a i the acceleration.

m=F/a

=4.719N/33 m/s²

=0.143 kg

5 0

3 years ago

In Figure 10.13, the author illustrates a book being raised at constant speed. Which of the following statements are true? Selec

sergey [27]

The true statement is that ;there is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.

We have to note that from the law of conservation of mechanical energy, the sum of the kinetic and potential energy of the book at any point remains constant.

If the book is being raised at constant speed, the the true statement is that ;there is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.

Learn more about kinetic energy: brainly.com/question/15308590

7 0

2 years ago

Read 2 more answers