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3 years ago

A car traveling 91 km/h is 280 m behind a truck traveling 76 km/h.

Physics

1 answer:

Olegator [25]3 years ago

6 0

Answer:

speed of car = 95 km/h

speed of truck = 75 km/h

relative speed of car with respect to truck = 95 - 75 = 20 km/h

now we will convert it into m/s

now time to cross the truck will be given as

time = 19.8 s

so it will take 19.8 s to cross the truck

Explanation:

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A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a

sladkih [1.3K]

Answer:

$d=1.29*10^{-6}m$

Explanation:

From the question we are told that:

Distance of wall from CD $D=1.4$

Second bright fringe $y_2= 0.803 m$

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

$y=frac{n*\lambda*D}{d}$

Where

$d=\frac{n*\lambda*D}{y}$

$d=\frac{2*431 *10^{-9}m*1.4}{0.803}$

$d=1.29*10^{-6}m$

8 0

2 years ago

What does the zigzag line in the circuit diagram represent

faust18 [17]

A) An electrical resistor

Hope that helps, Good luck! (:

6 0

3 years ago

A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori

yarga [219]

To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.

First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg = 1.55 * 9.81 
T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3

Tan θ = sin θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.

T then is equal to 20.20 N

4 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .