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3 years ago

11

A girl delivering newspapers travels one block west then three blocks north then five blocks east. what is her resultant displac

ement?

1 answer:

nlexa [21]3 years ago

8 0

3^2+4^2=25^2 = 5 units

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The elastic energy stored in your tendons can contribute up to 35% of your energy needs when running. Sports scientists have stu

Anni [7]

Answer:

ΔE = 9.7mJ

Explanation:

See attachment below.

7 0

3 years ago

The coefficient of linear expansion of ordinary glass is three times that of Pyrex glass. An ordinary glass rod and a Pyrex glas

DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length given as

ΔL = L α ΔT

Therefore

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}$

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}$

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

6 0

3 years ago

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

$v^2-u^2=2as$

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

5 0

2 years ago

A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo

guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}$

If we consider than the direction of the electric field is $\vec{E}=E_0\hat{x}$ , we can solve the problem differentiating the integral for each face of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6$

E₀ is a constant and each surface is equal to each other, so: $S_1=S_2=S_i=S$

Therefore:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c$

3 0

3 years ago

They are adding steel components to the roof supports of the structure. We know that steel elongates when heated. At what temper

Andru [333]

Answer:

The change in temperature of 576.9°C will produce an elongation of 9 inches per feet in steel.

Explanation:

The formula for linear expansion of a material is:

ΔL = αLΔT

where, ΔL = change in length

L = Original length

ΔT = Change in temperature

α = coefficient of linear expansion

For steel, α = 13 x 10^-6 /°C

L = 100 ft

ΔL = (9 in)(1 ft/12 in) = 0.75 ft

Therefore,

0.75 ft = (13 x 10^-6 /°C)(100 ft)ΔT

<u>ΔT = 576.9°C</u>

<u></u>

8 0

3 years ago