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lisov135 [29]
2 years ago
12

You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

-turn to go home, pick it up and now after 20s you are driving south with a speed of 20 m/s. What is your displacement after 20s?
Physics
1 answer:
Readme [11.4K]2 years ago
8 0

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

d_1 = v*t_1

d_1 = 10*20 = 200 m

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

d_2 = v*t = 20*20 = 400 m

so the net displacement is given as

\vec d = \vec d_2 - \vec d_1

\vec d = 400 - 200 = 200 m

so it will be displaced by total displacement 200 m

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A 1-kg mass is dropped from a third floor window. The acceleration of the mass is found to be 8 m/s2. What is the average force
Paha777 [63]
Summary:
m=1kg
a=8 m/s^2
g= 9,8 m/s^2
F(ar)=?


I hope to help you

7 0
2 years ago
A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

We need to calculate the velocity of the ball

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

Using conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

Put the value into the formula

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

d =vt

Where. d = distance

t = time

v = velocity

Put the value into the formula

d=2.922\times 0.587

d=1.72\ m

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0
2 years ago
A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38
oksano4ka [1.4K]

Answer:

the ratio is \frac{V_2}{V_1}=\sqrt{2}

Explanation:

Given

Initial Temperature T_1=387 KFinal Temperature T_2=774 K

The RMS velocity of molecules in a gas is given by

V_{rms}=\sqrt{\dfrac{3k_bT}{m}}

where T=temperature

k_b=constant

For T = 387K

V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1

For T = 774

V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)

dividing eqn 1 and eqn 2

\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}

\frac{V_2}{V_1}=\sqrt{2}

Thus,the ratio is \frac{V_2}{V_1}=\sqrt{2}

5 0
2 years ago
Read 2 more answers
A length of 20-gauge copper wire (of diameter 0.8118 mm) is formed into a circular loop with a radius of 30.0 cm. A magnetic fie
IrinaK [193]

Answer: 3 x 10^-24 watt

Explanation:

P ( resistivity) = 1.72e-8 (from the chart).

L= 2pi r

r= 30 cm.

R= pL/A

A= pi* r1^2

r1= 0.8118/2 * 10^-3 m

R= 1.68 x 10^-8 x (2x3.142x0.3)

= 3.24 x 10^-8

E=N do/dt

do= B* A

A= pi* 0.3^2

N=1

E = 1 x (14 x 3.142x 0.09) = 3.95

I=v/R

v=E,

I = 3.95 / 3.24 x 10^-8 = 1.22 x 10^8

P=I^2 x R.

= 3 x 10^-24 watt

7 0
1 year ago
What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i
Katarina [22]

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0
1 year ago
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