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3 years ago

An automobile travels on a straight road at 30 km/h How long will it take to travel 40 km

Physics

1 answer:

Anna [14]3 years ago

4 0

Answer:

the time taken for the automobile to travel the given distance is 80 mins.

Explanation:

Given;

speed of the automobile, v = 30 km/h

distance traveled by the automobile, x = 40 km

The time taken for the automobile to cover the given distance is calculated as;

time = distance / speed

time = 40 / 30

time = 1.3333 hours = 1.3333 hr x 60mins/hr = 80 mins

Therefore, the time taken for the automobile to travel the given distance is 80 mins.

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If a shot is put an angle of 41 degrees relative to the horizontal with a velocity of 36 ft/s in the direction of the put, what

Anarel [89]

Idk no idea none zero sorry hope someone know more than me

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3 years ago

An object with 274 J of GPE is 140cm above the ground. What is its mass?

Amanda [17]

E=274J
h=140cm=1,4m
g≈9,8m/s²

m=?

E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg

"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."

Regards M.Y.

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3 years ago

A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the

iris [78.8K]

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>

Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I

n = no. of turns per unit length and I = current through the solenoid

<h3>What's is the magnetic field inside the solenoid here?</h3>

Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

brainly.com/question/22814970

#SPJ4

6 0

1 year ago

Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41

alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 = 0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

3 0

3 years ago

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

5 0

3 years ago