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2 years ago

Why do the stars appear like pointed objects

Physics

1 answer:

saw5 [17]2 years ago

8 0

Answer:

because of where we are and our distance away from them, in reality they're giant balls of burning gas, like the sun. some stars are even bigger than the sun

Explanation:

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: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

The stoplights on a street are designed to keep traﬃc moving at 26 mi/h. the average length of a street block between traﬃc ligh

bonufazy [111]

We use the formula,

$v = \frac{d}{t}$ .

Here, v is velocity and its value given 26 mi/h ( in m/s, $\frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s$ ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

$t = \frac{80 m}{11.62m/s } = 6.88 \ s$

Thus, the time delay between green lights on successive blocks to keep the traﬃc moving continuously is 6.88 s

3 0

2 years ago

1. Is the relationship between velocity and centripetal force a direct, linear relationship or is it a nonlinear square relation

Svetllana [295]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

7 0

2 years ago

Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving

Iteru [2.4K]

Explanation:

Let $m_p$ is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

$\dfrac{mv^2}{r}=qvB$

r is the radius of path,

$r=\dfrac{mv}{qB}$

Time period is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi m_p}{qB}$

Frequency of proton is given by :

$f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}$

The wavelength of radiation is given by :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{2\pi m_pc}{qB}$

So, the wavelength of radiation produced by a proton is $\dfrac{2\pi m_pc}{qB}$ . Hence, this is the required solution.

3 0

2 years ago

The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

5 0

2 years ago

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Why do the stars appear like pointed objects​

Why do the stars appear like pointed objects