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2 years ago

What does a “fixed amount of fuel” mean?

Physics

2 answers:

klemol [59]2 years ago

6 0

Answer:

A "fixed price" energy tariff means that your unit price for gas and electricity will not change for the duration of the plan. A variable rate means your energy price can vary during the plan. If you have a fixed price plan and your supplier announces a price rise, your rates will not change.

Explanation: Google

Anna71 [15]2 years ago

5 0

Answer:

A "fixed price" energy tariff means that your unit price for gas and electricity will not change for the duration of the plan. A variable rate means your energy price can vary during the plan

A very large amount

Explanation:

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The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end

Darina [25.2K]

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

$T = mg$

here we will have

$T = 2(9.8) = 19.6 N$

now we will have speed of wave given as

$v = \sqrt{\frac{T}{\mu}}$

here we will have

$v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}$

$v = 16.33 m/s$

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

$\lambda = \frac{v}{F}$

$\lambda = \frac{16.33}{100} = 0.16 m$

so wavelength will be 0.16 m

5 0

3 years ago

An orchestra is practicing a piece that is to be played at an allegro tempo. The conductor sets a metronome at 160 beats per min

omeli [17]

Answer:

375 ms

Explanation:

the frequency of metronome , f = 160 beats per minute

f = 160 /60 beats per sec

f = 2.67 beats /s

the period of a single beat , T = 1/f

T = 1/2.67 s

T = 0.375 s = 375 ms

the period of a single beat is 375 ms

3 0

3 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

3 years ago

With what force will a car hit a tree if the car has a mass of 3,415 kg and it is accelerating at

weeeeeb [17]

Answer:

$F= 17,075\ N$

Explanation:

When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.

The accelerating force can be calculated using Newton's second law:

$F=m\cdot a$

Where m is the mass of the car and a is the acceleration.

$F=3,415\ kg\cdot 5\ m/s^2$

$\boxed{F= 17,075\ N}$

3 0

3 years ago

A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl

guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

$P_1 = m_av_a + m_b v_b$

$P_1 = 1300 \times 35+ 1000 \times 25$

$P_1 =70500 Kg.m/s$

final momentum

$P_2 = m_aV_a + m_b V_b$

$P_2 = 1300 \times 30+ 1000 \times 31.5$

$P_2 =70500 Kg.m/s$

here initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0

3 years ago