The speed of a wave on a rope is 50cm/s an its wave length is 10cm. What is its frequency

What is the momentum of a 10kg ball moving at a velocity of 4m/s east?

Katyanochek1 [597]

Answer:

40 kg.m/s

Explanation:

Momentum, p is defined as the product of mass and velocity of an object. Numerically, it is represented as, p=mv where m is mass of the object and v is the velocity in which the object moves, with keen observation on the direction before and after collision. Substituting 10 kg for m and 4 m/s for v then momentum, P=10*4=40 kg.m/s

6 0

3 years ago

While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile's acceleration? (Reme

slava [35]

Answer:

Acceleration = -0.75 m/s²

Explanation:

Given the following data;

Initial velocity = 24 m/s

Final velocity = 15 m/s

Time = 12 seconds

To find the acceleration of the automobile;

Acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac {final \; velocity - initial \; velocity}{time}$

Substituting into the formula, we have;

$Acceleration = \frac{15 - 24}{12}$

$Acceleration = \frac{-9}{12}$

Acceleration = -0.75 m/s²

Therefore, the automobile is decelerating because its final velocity is lesser than its initial velocity, leading to a negative value.

5 0

2 years ago

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to

uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0

2 years ago

A truck going 15 kmôŹ°€h has a head-on collision with a small car going 30 kmôŹ°€h. Which statement best describes the situation

zlopas [31]

1. e) None of the above is necessarily true.

2.d) Without knowing the mass of the boat and the sack, we cannot tell.

5 0

3 years ago

Add the vectors:

Anettt [7]

Vector 1 has components

$x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m$

$y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m$

and vector 2 has

$x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m$

$y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m$

Add these vectors to get the resultant, which has components

$x_{\rm total}\approx11.133\,\mathrm m$

$y_{\rm total}\approx13.268\,\mathrm m$

The magnitude of the resultant is

$\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m$

with direction $\theta$ such that

$\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ$

or about 50º N of E.

8 0

3 years ago