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2 years ago

John runs 1.0 m/s at first, and then accelerates to 1.6 m/s during

Physics

2 answers:

erastova [34]2 years ago

7 0

Answer: $0.13m/s^2$

Explanation:

$Formula: a=\frac{V_2-V_1}{t}$

Where;

a = acceleration

V2 = final velocity

V1 = initial velocity

t = time

If John runs 1.0 m/s first, we assume this is V1. He accelerates to 1.6 m/s; this is V2.

$a=\frac{1.6m/s-1.0m/s}{4.5s}$

$a=\frac{0.6m/s}{4.5s}$

$a=0.13m/s^2$

irakobra [83]2 years ago

4 0

Answer:.13

Explanation:

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A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at

GuDViN [60]

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K

4 0

2 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?

barxatty [35]

Answer:

$2.5\cdot 10^{-4}$

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

$S=\frac{\Delta L}{L_0}$

where

$\Delta L$ is the change in length of the object

$L_0$ is the original length of the object

In this problem, we have $L_0 = 4 m$ and $\Delta L=1 mm=0.001 m$ , therefore the strain is

$S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}$

5 0

2 years ago

When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pull

Mandarinka [93]

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

$F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}$

The force of the sun is, therefore, $10^{10}$ times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

5 0

1 year ago

When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What

ser-zykov [4K]

Answer:

Electric field, $E=4\times 10^5\ N/C$

Explanation:

It is given that,

Magnitude of charge, $q_o=2\ nC=2\times 10^{-9}\ C$

Force experienced, $F=8\times 10^{-4}\ N$

We need to find the electric field at the origin. It is given by :

$F=q_o\times E$

$E=\dfrac{F}{q_o}$

$E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}$

$E=4\times 10^5\ N/C$

So, the electric field at the origin is $4\times 10^5\ N/C$ . Hence, this is the required solution.

3 0

2 years ago