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3 years ago

13

A 3kg object moving at 4 m/s encounters a 20 N resistive force over a duration of 0.20 seconds. The impulse experienced by this

object is _____

1 answer:

WITCHER [35]3 years ago

3 0

Answer:

the impulse experienced by this object is 4 Ns

Explanation:

Given;

mass of the object, m = 3 kg

velocity of the object, v = 4 m/s

resistive force, F = 20 N

duration of impact, t = 0.2 s

The impulse experienced by this object is calculated as follows;

J = F x t

J = 20 x 0.2

J = 4 Ns

Therefore, the impulse experienced by this object is 4 Ns

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A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at

riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

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2 years ago

the attendance qt a seminar in a year 1 was 500 in year 2 the attendance changed by 100 what was the percent change in attendanc

TiliK225 [7]

20% change from year one to year two.

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3 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

2 years ago

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

Oksana_A [137]

Answer:

a) $\omega=1.36rad/s$

b) $\omega=12.99rpm$

c) $F=705.6N$

Explanation:

a) The angular velocity is related to the centripetal acceleration by the formula $a_{cp}=\omega^2r$ , which for our purposes we will write as:

$\omega=\sqrt{\frac{a_{cp}}{r}}$

Since <em>we want this acceleration to be 1.5 times that due to gravity</em>, for our values we will have:

$\omega=\sqrt{\frac{1.5g}{r}}=\sqrt{\frac{(1.5)(9.8m/s^2)}{(8m)}}=1.36rad/s$

b) 1 rpm (revolution per minute) is equivalent to an angle of $2\pi$ radians in 60 seconds:

$1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s$

Which means <em>we can use the conversion factor</em>:

$\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1$

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

$\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm$

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

$F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N$

8 0

2 years ago

A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order t

NikAS [45]

Answer:

C. $\frac{25}{7}$ m

Explanation:

We are given that

Weight of board=w=10 N

Length of board=L=5 m

Tension in the string=T=3 N

Applied upward force=F=7 N

We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

The board is uniform therefore, the center of board is the mid- point of board.

Therefore, the lever arm of weight= $r_1=\frac{L}{2}=\frac{5}{2}m$

Now, the torque exerted by the weight of the board

$\tau_1=Force\times perpendicular\;distance=10\times \frac{5}{2}=25 N$

The torque exerted by applied force= $\tau_2=7\times r=7r$

In static equilibrium

The sum of rotational forces=0

$\tau_1+\tau_2=0$

The two rotational force act in opposite direction therefore,

$\tau_2=-7r$

Substitute the values

$25-7r=0$

$7r=25$

$r=\frac{25}{7}$ m

Hence, option C is true.

7 0

2 years ago