Which statement is true of convex mirrors?

At a certain time a particle had a speed of 48 m/s in the positive x direction, and 4.5 s later its speed was 92 m/s in the oppo

larisa86 [58]

Answer:

$-31.1 m/s^2$

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is calculated as:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it is important to also take into account the direction of the velocity.

For the particle in this problem, we have:

u = +48 m/s is the initial velocity (positive direction)

v = -92 m/s is the final velocity (negative direction)

t = 4.5 s is the time interval

Therefore, the average acceleration is

$a=\frac{v-u}{t}=\frac{-92-(+48)}{4.5}=-31.1 m/s^2$

4 0

3 years ago

A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc

balandron [24]

Answer:

Approximately $28^{\circ}$ .

Explanation:

The refractive index of the air $n_{\text{air}}$ is approximately $1.00$ .

Let $n_\text{glass}$ denote the refractive index of the glass block, and let $\theta _{\text{glass}}$ denote the angle of refraction in the glass. Let $\theta_\text{air}$ denote the angle at which the light enters the glass block from the air.

By Snell's Law:

$n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})$ .

Rearrange the Snell's Law equation to obtain:

$\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}$ .

Hence:

$\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}$ .

In other words, the angle of refraction in the glass would be approximately $28^{\circ}$ .

7 0

2 years ago

Which of the following is a type of T cell?

Bess [88]

Answer:

D

Explanation:

7 0

3 years ago

Read 2 more answers

Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed

photoshop1234 [79]

Answer:

The railroad tracks are 13 m above the windshield (12 m without intermediate rounding).

Explanation:

First, let´s calculate the time it took the driver to travel the 27 m to the point of impact.

The equation for the position of the car is:

x = v · t

Where

x = position at time t

v = velocity

t = time

x = v · t

27 m = 17 m/s · t

27 m / 17 m/s = t

t = 1.6 s

Now let´s calculate the distance traveled by the bolt in that time. Let´s place the origin of the frame of reference at the height of the windshield:

The position of the bolt will be:

y = y0 + 1/2 · g · t²

Where

y = height of the bolt at time t

y0 = initial height of the bolt

g = acceleration due to gravity

t = time

Since the origin of the frame of reference is located at the windshield, at time 1.6 s the height of the bolt will be 0 m (impact on the windshield). Then, we can calculate the initial height of the bolt which is the height of the railroad tracks above the windshield:

y = y0 + 1/2 · g · t²

0 = y0 -1/2 · 9.8 m/s² · (1.6 s)²

y0 = 13 m

8 0

3 years ago

A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car

jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

8 0

2 years ago