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2 years ago

If a object is at rest, does it mean there are no forces acting on it

Physics

2 answers:

kotegsom [21]2 years ago

5 0

Yessssssss that is correct and the reason why is because the force

Ganezh [65]2 years ago

4 0

Yes that is correct :)

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9) What will happen to the period of a pendulum if we change the rope of a pendulum with another which is four times with the in

Arturiano [62]

Double

Explanation:

Since the period T of a pendulum is given by

$T = 2\pi \sqrt{\dfrac{l}{g}}$

By increasing the length of the pendulum by 4, the period becomes

$T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T$

You can see that the period doubles when we increase the length by a factor of 4.

5 0

2 years ago

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m

Gelneren [198K]

Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

= 4.98 × 10¹⁹ protons

≅ 5 × 10¹⁹ protons

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

= 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

= 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

= 30 × 10⁷/14.62

= 2.05 × 10⁷ °C

5 0

2 years ago

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference

strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

$n_{air}=1$

$\theta_{liquid} = 19.38\°$

$\theta_{air}35.09\°$

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

$n_1sin\theta_1 = n_2sin\theta_2$

$n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}$

$n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}$

Replacing the values we have:

$n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}$

$n_liquid = 1.7323$

Therefore the refractive index for the liquid is 1.7323

6 0

3 years ago

What is the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s?

Alik [6]

1/2mv^2

1/2x12x10^2=600J

The kinetic energy is 600J

3 0

3 years ago

Read 2 more answers

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

8 0

3 years ago