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4 years ago

Which answer accurately describes the difference between speed and velocity

Physics

2 answers:

tresset_1 [31]4 years ago

8 0

Answer:

Two answers C and D

Explanation:

Because they are both the same and describe this well

Usimov [2.4K]4 years ago

3 0

I think the answer is D but Check itt

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Astronomers use radio waves to learn about the universe. Radio waves travel light-years through space before reaching Earth. Rad

IrinaVladis [17]

Electromagnetic wave, a transverse wave

4 0

4 years ago

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2

Svetradugi [14.3K]

Answer:

B. to the right

Explanation:

Given:

test charge +q

distance of the test charge from +Q, r

distance of test charge from +2Q, 2r

<u>Force on the test charge due to +Q:</u>

$F_1=k.\frac{Q.q}{r^2}$

<u>Force on the test charge due to +Q:</u>

$F_2=k.\frac{2Q.q}{(2r)^2}$

$F_2=k.\frac{Q.q}{2r^2}$

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

3 0

3 years ago

Read 2 more answers

Problem Page Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 28. g o

sergejj [24]

Answer : The maximum mass of carbon dioxide produced by the chemical reaction can be, 82.3 grams.

Solution : Given,

Mass of $C_2H_6$ = 28 g

Mass of $O_2$ = 190 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $C_2H_6$ and $O_2$ .

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{28g}{30g/mole}=0.933moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{190g}{32g/mole}=5.94moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

From the balanced reaction we conclude that

As, 2 mole of $C_2H_6$ react with 7 mole of $O_2$

So, 0.933 moles of $C_2H_6$ react with $\frac{7}{2}\times 0.933=3.26$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_6$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $C_2H_6$ react to give 4 mole of $CO_2$

So, 0.933 moles of $C_2H_6$ react to give $\frac{4}{2}\times 0.933=1.87$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(1.87moles)\times (44g/mole)=82.3g$

Therefore, the maximum mass of carbon dioxide produced by the chemical reaction can be, 82.3 grams.

6 0

4 years ago

A uniform magnetic field of magnitude 0.72 T is directed perpendicular to the plane of a rectangular loop having dimensions 8.2

Liono4ka [1.6K]

Answer:

Explanation:

Magnetic flux is expressed as the product of magnetic field and cross sectional area.

Φ = BAsintheta

Given

B = 0.72T

A = 8.2cm×14cm

A = 0.082m × 0.14m

Area = 0.01148m²

Theta = 90°

Substitute into the formula

Φ = BAsintheta

Φ = 0.72(0.01148)sin90°

Φ = 0.72(0.01148)(1)

Φ = 0.0082656

Hence the magnetic flux through the loop is 8.2656 × 10^-3 Weber

8 0

3 years ago

A proton in a uniform electric field moves along a straight line with constant acceleration. Starting from rest it attains a vel

Sav [38]

a) The acceleration of the proton is $5.0\cdot 10^{13} m/s^2$

b) The time required to reach the given velocity is $2\cdot 10^{-8}s$

Explanation:

This is a motion at constant acceleration, so we can use the following suvat equation:

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

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the proton in this problem, we have:

$v=1,000,000 m/s$ is the final velocity

$u=0$ is the initial velocity (it starts from rest)

$s = 0.01 m$ is the distance covered

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{(1,000,000)^2-0}{2(0.01)}=5.0\cdot 10^{13} m/s^2$

For this part, we can use the following suvat equation instead:

$v=u+at$

where:

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken for the velocity to change from u to v

We have here the following data:

$v=1,000,000 m/s$ is the final velocity

$u=0$ is the initial velocity (it starts from rest)

$a=5.0\cdot 10^{13} m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{1,000,000}{5.0\cdot 10^{13}}=2\cdot 10^{-8}s$

Learn more about accelerated motion here:

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5 0

4 years ago