B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?

irakobra [83]

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

Original Temperature = 280K

New Temperature = 700K

To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$VT = K$

$\frac{V1}{T1} = \frac{V2}{T2}$

Making V2 as the subject formula, we have;

$V_{2}= \frac{V1}{T1} * T_{2}$

$V_{2}= \frac{2}{700} * 280$

$V_{2}= 0.0029 * 280$

V2 = 0.8L

Therefore, the volume of the gas after it is heated is 0.8L.

7 0

3 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

3 years ago

An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

Leya [2.2K]

Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
$F=qvB \sin \theta$
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
$\theta$ the angle between the directions of v and B.

Re-arranging the formula, we find:
$v= \frac{F}{qB \sin \theta}$
and by substituting the data of the problem (the charge of the electron is $q=1.6 \cdot 10^{-19} C$ ), we find the velocity of the electron:
$v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s$

4 0

3 years ago

Why are lunar calendars not more widely used today?

Nookie1986 [14]

The seasons affect the scheduling of far more events the moon phases do.

6 0

3 years ago

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