Calculate 2 x 10 -3 cm ÷ 2.5 x 10 4 cm

Write the formular of force experienced by charge in electric field

lesya [120]

The answer is b) force exponerse by charge in electric field

7 0

3 years ago

Read 2 more answers

When going from a fast speed to a slow speed how is light bent?

Zolol [24]

Toward the normal line

4 0

3 years ago

Read 2 more answers

Shane is riding his bike at a speed of 10m/s. He isn't watching carefully where he is going and almost hit a tree. He abruptl

Rom4ik [11]

Answer:

2.5m/s²

Explanation:

a = v/t

Where;

V = velocity (m/s)

a = acceleration (m/s²)

t = time (s).

According to the information provided in this question,

a = ?

v = 10m/s

t = 4

a = 10/4

a = 2.5m/s²

4 0

3 years ago

An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach

Gelneren [198K]

Answer: $5 m/s^{2}$

Explanation:

This problem is related to vertical motion, and the equation that models it is:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

5 0

3 years ago

A 900 kg vehicle moves around a curve with an incline of 20\circ∘ at a speed of 12.5 m/s. If the curve has a radius of 50 meters

valentina_108 [34]

Answer:

The normal force experienced by the car is approximately 8223.2 N

Explanation:

The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction

The mass of the vehicle that moves around the curve, m = 900 kg

The incline of the curve, θ = 20°

The speed with which the vehicle moves around the curve, v = 12.5 m/s

The radius of the curve, R = 50 meters

We have;

$N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}$

Where;

θ = The angle of inclination of the road = 20°

N = The normal force experienced by the car

m = The mass of the car = 900 kg

v = The velocity with which the car is moving = 12.5 m/s

R = The radius of the curve around which the vehicle moves = 50 m

$\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})} = 8223.1998754586828969046217875927$

The normal force experienced by the car = N ≈ 8223.2 N.

6 0

3 years ago