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3 years ago

A boat moves initially at 10 m/s, then accelerates at 2 m/s^2 for 10 s. What is the velocity of the boat after 10 s?

Physics

1 answer:

Fudgin [204]3 years ago

6 0

Answer:

30 m/s

Explanation:

A boat moves initially at 10 m/s

The boat accelerates at 2 m/s^2 for 10 seconds

We are required to calculate the final velocity (v)

Therefore the velocity of the boat can be calculated as follows

v = u + at

u= 10 m/s

a= 2 m/s^2

t= 10 seconds

v= 10 + 2(10)

v= 10 + 20

= 30 m/s

Hence the velocity of the boat after 10 seconds is 30 m/s

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Ce unitate de masura are indicele de REFRACTIE?

ivolga24 [154]

Answer:se obesrva ca indicele de refractie NU are unitate de masura, este adimensional. Se exprima print-un raport, de exemplu 4/3, 1,5 etc.

Explanation:

3 0

3 years ago

) A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the f

Yuri [45]

The frictional force required is 9000 N

Explanation:

In order to keep the car in the turn in circular motion without sliding, the frictional force must provide the centripetal force necessary for the circular motion.

Therefore, we can write:

$F_f = m \frac{v^2}{r}$

where the term on the left is the frictional force while the term on the right is the centripetal force, and where:

m is the mass of the car

v is its speed

r is the radius of the curve

For the car in this turn, we have

m = 1000 kg

v = 30 m/s

$r=\frac{0.20 km}{2}=0.10 km = 100 m$ (since the diameter is 0.20 km, the radius is half that value)

And substituting, we find

$F_f = (1000) \frac{30^2}{100}=9000 N$

Learn more about friction:

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8 0

4 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25

torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{V}{d}$

Put the value into the formula

$E=\dfrac{25.0}{1.50\times10^{-3}}$

$E=16666.66\ V/m$

We need to calculate the charge density

Using formula of charge density

$\sigma=E\times\epsilon_{0}$

Put the value into the formula

$\sigma=16666.66\times8.85\times10^{-12}$

$\sigma=1.474\times10^{-7}\ C/m^2$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}$

$C=4.484\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=4.484\times10^{-12}\times25.0$

$q=112.1\times10^{-12}\ C$

Hence, The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

8 0

3 years ago

A snail at position 3 cm moves to position 20 cm in 8 seconds.

natulia [17]

Answer: 17cm.

Explanation:

The equation you're using is:

Δd = df - di

Which means the change in position is equal to the final position minus the starting position. In this case that works out to 20cm - 3cm = 17cm. We're only interested in how much the snail moved, not how long it took to move, so even though they give a time it actually doesn't matter for this question.

4 0

4 years ago

At what point is the KINETIC ENERGY at its highest?

igomit [66]

Answer:

An object has the MOST kinetic energy when it's movement is the GREATEST.

Explanation:

8 0

3 years ago