What is the kinetic energy of a 2000 kilogram boat moving at 5m/sec?

An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr

AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

$a = \dfrac{400}{2\times 50}$

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

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

$\mu = \dfrac{4}{9.81}$

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0

2 years ago

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago

An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th

maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

3 0

2 years ago

Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value.

wolverine [178]

Answer:

V

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

$|I|=m(v-u)=(1000)(0-25)=25,000 Ns$

(II)

u = 25 m/s

v = 0

$|I|=m(v-u)=(1000)(0-25)=25,000 Ns$

(III)

In this case,

F = 2000 N is the force

$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

(IV)

F = 2000 N is the force

$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

(V)

u = 25 m/s

v = -25 m/s

$|I|=m(v-u)=(1000)(-25-25)=50,000 Ns$

So the ranking from largest to smallest is:

V

I and II

III and IV

5 0

2 years ago

If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

zalisa [80]

If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.

6 0

2 years ago

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