How many seconds will it take for a satellite to travel 450,000 m at a rate of 120 m/s?

What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a<br> velocity of 3.0 m/s?

Sav [38]

46.6666 that is the mass number

Explanation:

14 divided 3.0

5 0

3 years ago

A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal

Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

$v_x = v_i cos \theta$

where

$v_i = 40 m/s$ is the initial velocity

$\theta=20^{\circ}$ is the angle of projection

Substituting,

$v_x = (40)(cos 20^{\circ})=37.6 m/s$

And therefore, the range of the projectile is:

$d=v_x t = (37.6)(5.0)=188 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

A truck is moving at a constant speed of 50 m/s. A boy riding in the back of the truck throws a newspaper out the back of the tr

lbvjy [14]

Answer:

70m/s

Explanation:

7 0

4 years ago

A car is traveling in uniform circular motion on a section of flat roadway whose radius is . The road is slippery, and the car i

Alona [7]

Answer:

The smallest radius will be four (4) times the initial radius

Explanation:

The car maintains a constant angular speed. According to Newton's Second Law F = m a

1. $F_{r}=m*A_{n}$

2. $A_{n}=\frac{v^2}{R_{p}}$

Replacing 2 in 1

3. $F_{r}=m*\frac{v^2}{R_{p}}$

Where:

Fr= Frictional force

Rp= Initial Radius

An= Centripetal Acceleration

M= Mass

V= Velocity

Also we have that:

4. $F_{r}=\mu *W=\mu*m*g$

μ= Coefficient of friction between the car and the surface

M= Mass

W= Weight

G= Gravity

r is cleared from equation 3

5. $R_{p}=m*\frac{v^2}{F_{r}}$

Replacing 4 in 5

6. $R_{p}=m*\frac{v^2}{\mu*m*g}$

Simplifying

7. $R_{p}=\frac{v^2}{\mu*g}$

Now we have a new velocity equal to twice the initial velocity, We replace it by 2v in equation 7

8. $R_{n}=\frac{(2v)^2}{\mu*g}$

Computing

9. $R_{n}=\frac{4v^2}{\mu*g}$

Replacing 5 in 9

$R_{n}=4*R_{p}$

8 0

3 years ago

Two charged spheres are 20cm apart and exert an attractive force of 6x10^-9 N on each other. What will be the force of attractio

Travka [436]

The force of attraction between 2 charged spheres can be explained by Coulomb's law,
It states the force of attraction is directly proportional to the magnitudes of the charges and inversely proportional to the square of the distance between the charges.

/ $F = \frac{k q_{1} q_{2} }{ r^{2} }$
where F - force of attraction/repulsion
q₁ and q₂ - charges of the 2 spheres
k - Coulomb's law constant
r - distance between the spheres

In the question given, the charges of the spheres remain constant in both instances, only distance changes. Therefore (kq₁q₂) = c which is a constant
then F = c / r²
first instance
6 x 10⁻⁹ N = c/ (20 cm)² ---1)
F = c/(10 cm)² --- 2)
2) / 1)
$\frac{F}{6* 10^{-9} } = \frac{400}{100}$
F = 6 x 10⁻⁹ x 4
F = 2.4 x 10⁻⁸ N

8 0

3 years ago