How do mass and velocity of an object affects mommentum?

At a sports car rally, a car starting from rest accelerates uniformly at a rate of 5 m/s/s over a straight-line distance of 291

Bogdan [553]

Answer:

10.8 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s/s

Distance travelled (s) = 291 m

Time (t) taken =?

We can calculate the time taken for the car to cover the distance as follow:

s = ut + ½at²

291 = 0 × t + ½ × 5 × t²

291 = 0 + 2.5 × t²

291 = 2.5 × t²

Divide both side by 2.5

t² = 291 / 2.5

t² = 116.4

Take the square root of both side

t = √116.4

t = 10.8 s

Thus, it will take the car 10.8 s to cover the distance.

8 0

3 years ago

What measurements will be made to determine the magnitude of the test-mass centripetal acceleration?

DedPeter [7]

Answer:period, spring constant, radius of circular part, velocity of the test mass, mass of the test-mass, mass of the hanging mass

Explanation:

3 0

4 years ago

Which is an example of transforming potential energy to kinetic energy? Select two options

blsea [12.9K]

Answer : 

.changing chemical energy to thermal energy

.changing nuclear energy to radiant energy

Explanation : 

<h3>Hope it helps you mate ))</h3>

3 0

3 years ago

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

Free_Kalibri [48]

Answer:

$\beta = 41.68°$

Explanation:

according to snell's law

$\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }$

refractive index of water n_w is 1.33

refractive index of glass n_g is 1.5

$sin\alpha = \frac{n_w}{n_g}* sin30$

$sin\alpha = 0.443$

now applying snell's law between air and glass, so we have

$\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}$

$sin\beta = \frac{n_g}{n_a} sin\alpha$

$\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]$

we know that $sin\alpha = 0.443$

$\beta = 41.68°$

7 0

3 years ago

) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)

natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

$\tau = \vec{F} \times \vec{r}$

$\tau = (8i+6j)\times(-3i+4j)$

$\tau = (8*4)(i\times j)+(6*-3)(j\times i)$

$\tau = 32k +18k$

$\tau = 50 k$

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

$cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}$

$cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}$

$cos\theta = -0.24$

$\theta = cos^{-1} (-0.24)$

$\theta = 103.88\°$

Therefore the angle between the ratio and the force is 103.88°

5 0

3 years ago