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2 years ago

A car travels 10 miles East in 30 minutes. What is its velocity in miles per hour?

Physics

2 answers:

Stella [2.4K]2 years ago

8 0

Answer:

Velocity = 20 mph

Explanation:

Distance = 10 miles

Time = 30 min = 0.5 hour

Velocity = distance/time

= 10miles/0.5 hour

= 20 mph

Ghella [55]2 years ago

7 0

The car's velocity is

20 miles per hour East.

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A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

KIM [24]

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

5 0

3 years ago

Read 2 more answers

Describe how the chemical bonds changed during the reaction.

k0ka [10]

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Answer:

In a chemical reaction, the atoms and molecules that interact with each other are called reactants. ... No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

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6 0

3 years ago

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If an X-ray tube is operating at a current of 30.0 mA:a) How many electrons are striking the target per second? b) If the potent

Zina [86]

Answer:

Explanation:

A. Q= It

30E-3A=q

Ne = q

30E-3/1.6*10-19= N

N= 1.8*10^16 electrons

B. Power= I x v

= 30*10^-3A x 100*10^3v

= 3000watts

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3 years ago

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.50 T. The field is then adjusted so that a

bixtya [17]

Answer:

$B_e = 2.72 \times 10^{-4} T$

Explanation:

As we know that when charge particle is projected in perpendicular magnetic field then the radius of the charge particle is given as

$F = qvB$

$\frac{mv^2}{r} = qvB$

now we have

$r = \frac{mv}{qB}$

since here radius of proton and electron will be same

so we will have

$r_e = r_p$

$\frac{m_e v}{q_e B_e} = \frac{m_p v}{q_p B_p}$

so we have

$B_e = \frac{B_p m_e}{m_p}$

given that

$B_p = 0.50 T$

$m_e = 9.11 \times 10^{-31} kg$

$m_p = 1.67 \times 10^{-27} kg$

so we have

$B_e = \frac{0.50(9.11\times 10^{-31})}{1.67\times 10^{-27}}$

$B_e = 2.72 \times 10^{-4} T$

4 0

3 years ago

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How much heat is required to convert 2.55g of water at 28 degrees c to steam?

pantera1 [17]

There are two different processes here:
1) we must add heat in order to bring the temperature of the water from $28^{\circ}C$ to $100^{\circ}C$ (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
$Q_1=m C_s \Delta T$
where
$m=2.55 g$ is the water mass
$C_s = 4.18 g/J^{\circ}C$ is the water specific heat
$\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C$ is the variation of temperature of the water
If we plug the numbers into the equation, we find
$Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J$

2) The heat needed for process 2) is
$Q=m L_e$
where
$m=2.55 g$ is the water mass
$L_e = 2264.7 J/g$ is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
$Q_2=(2.55 g)(2264.7 J/g)=5775.0 J$

So, the total heat needed for the whole process is
$Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J$

4 0

3 years ago