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2 years ago

If a car with an initial velocity of 10 m/s accelerates at a rate of 50 m/s^2

Physics

1 answer:

expeople1 [14]2 years ago

6 0

Given:

Initial velocity (u) = 10 m/s

Acceleration (a) = 50 m/s²

Time taken (t) = 3 s

To Find:

Final velocity (v)

Answer:

By using equation of motion, we get:

$\bf \longrightarrow v = u + at \\ \\ \rm \longrightarrow v = 10 + 50 \times 3 \\ \\ \rm \longrightarrow v = 10 + 150 \\ \\ \rm \longrightarrow v = 160 \: m {s}^{ - 1}$

$\therefore$ Final velocity (v) = 160 m/s

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A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the

Iteru [2.4K]

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

6 0

3 years ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it

inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

$F_{b}=\rho_{air}\times V_{b}\times g$

Where, $\rho_{air}$ = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

$F_{b}=1.20\times321\times9.81$

$F_{b}=3778.8\ N$

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

A ball is thrown at an angle of 38° to the horizontal. What happens to the

finlep [7]

Answer:

Vy = V0 sin 38 where Vy is the initial vertical velocity

The ball will accelerate downwards (until it lands)

Note the signs involved if Vy is positive then g must be negative

The acceleration is constant until the ball lands

t (upwards) = (0 - Vy) / -g = Vy / g final velocity = 0

t(downwards = (-Vy - 0) / -g = Vy / g final velocity = -Vy

time upwards = time downwards (conservation laws)

8 0

3 years ago

10.0 ppm 10.4 ppm 10.2 ppm 10.8 ppm 10.1 ppm 10.5 ppm 10.1 ppm Using the new instrument on the first day the technicians got a v

iren [92.7K]

Answer:

The datapoint 9.0 ppm is outlier at the 90% confidence level.

Explanation:

The old data has following values

mean=10.5 mm

standard deviation 0.2 mm

Now the mean of new values is calculated as following

$mean=\frac{10+10.4+10.2+10.8+10.1+10.5+10.1}{7}\\mean=10.3 ppm$

So the value as 9.0 ppm can be considered easily as outlier in this regard.

3 0

3 years ago