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3 years ago

5

Heather and Matthew walk with an average velocity of 0.98 m/s eastward. If it takes them 34 min to walk to the store, what is th

eir displacement?

1 answer:

TEA [102]3 years ago

3 0

Answer:

D= 1999.2 m

Explanation:

Given that

Average velocity ,v= 0.98 m/s

time ,t= 34 min

We know that

1 min = 60 s

That is why

t= 34 x 60 =2040 s

We know that

Displacement = Average velocity x time

D= v t

Now by putting the values in the above equation

D= 0.98 x 2040 m

D= 1999.2 m (eastward)

The direction of the displacement will be towards eastward.

That is why the displacement will be 1999.2 m or we can say that 1.9992 km.

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Draw the vector c⃗ =1.5a⃗ −3b⃗

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<span>The magnitude of a is 1.5
The magnitude of b is -3
The magnitude of the vector is
</span>√(1.5² + (-3)²) = 3.35
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</span>θ = tan⁻¹ (-3/1.5) = 63.43°
<span>The vector is drawn with a magnitude of 3.35 and an angle of 63.43</span>°.

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3 years ago

The frequency of an electromagnetic wave is 1E5 Hz. What is the wavelength of the light in meters?

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Just use the formula
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3 years ago

what is the acceleration produced by the resultant force acting on an object if the coefficient of friction acting between the b

LenKa [72]

The acceleration of the block is $-0.98 m/s^2$

Explanation:

The expression for the force of friction acting on the block is (assuming the surface is horizontal and flat):

$F_f = -\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the block

g is the acceleration of gravity

and where the negative sign means the direction of the force is opposite to that of the motion of the block

If this is the only force acting on the object, then this is also the resultant force, so we can rewrite Newton's second law as:

$F=ma\\-\mu mg = ma$

where

a is the acceleration of the block

Re-arranging the equation,

$a=-\mu g$

And so this is the expression for the acceleration of the block acted upon the force of friction.

In this problem, we have:

$\mu=0.1$

$g=9.8 m/s^2$

Solving,

$a=-(0.1)(9.8)=-0.98 m/s^2$

Learn more about friction:

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3 years ago

A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2 /3. Suppose a 2.00-m r

Stels [109]

To solve this problem we will apply the concepts related to the moment of inertia and Torque, the latter both its translational and rotational expression.

According to the information given the moment of inertia of the body would be

$I = \frac{1}{3} mL^2$

Replacing we have

$I = \frac{1}{3} (3kg)(2m)^2$

$I = 4 kg * m^2$

Now the translational torque would be the product between the force applied (Its own Weight) and the distance (Its center of mass at the middle)

$\tau = F*r$

$\tau = mg (\frac{L}{2})$

$\tau = (3)(9.8)(\frac{2}{2})$

$\tau = 29.4N\cdot m$

Now the rotational torque is defined as the product between the moment of inertia and the angular acceleration, then,

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

Replacing,

$\alpha = \frac{29.4}{4}$

$\alpha =7.35 rad / s^2$

Therefore the angular acceleration is $7.35rad/s^2$

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3 years ago

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

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3 years ago