The standard unit of brightness is called the candela.<br> True<br> False

o-na [289]

Explanation:

Let the distance covered by the body be s, initial and final velocities be u and v respectively and time taken be t.

$\therefore Average\: velocity = \frac{u+v}{2} \\\\Now, \:we \:know\: that\\\\Distance \:covered\\ = Average\: velocity \times time\\\\\therefore s= \frac{(u+v) }{2} \times t..... (1)\\\\$

By first equation of motion:

$v = u + at$

Substituting the value of v in equation (1), we find:

$s= \frac{(u+u + at)}{2} \times t\\\\\therefore s= \frac{(2u + at)}{2} \times t\\\\\therefore s= \frac{(2ut + at^2)}{2}\\\\\therefore s= \frac{2ut} {2}+ \frac{at^2}{2}\\\\ \huge \orange {\boxed {\therefore s= ut+ \frac{1}{2}at^2}} \\\\$

Hence proved.

6 0

3 years ago

In a circus performance, a monkey on a sled is given an initial speed of 4.3 m/s up a 24◦ incline. The combined mass of the monk

Serggg [28]

Answer:

d = 1.15 m

Explanation:

In absence of friction, the change in kinetic energy of the combined mass of the monkey and the sled, must be equal (with opposite sign), to the change in gravitational potential energy:

ΔK = -ΔU

When friction is not negligible, the change in mechanical energy, must be equal to the work done by non-conservative forces (kinetic friction in this case):

ΔK + ΔU = Wnc (1)

As the monkey + sled reach to the maximum distance up the incline, they will come momentarily to a stop, so the final kinetic energy is 0.

$(K_{f} -K_{o}) = 0 - \frac{1}{2} * m*v_{o} ^{2} = -\frac{1}{2} *22.5kg*(4.3m/s)^{2} = -208.1J$

The change in gravitational energy, can be written as follows:

$(U_{f} - U_{o} ) = m*g*h - 0 = m*g*h = \\ \\ 22.5 kg*9.8 m/s2*d*sin (24 deg) = 89.7J*d$

The sum of these two quantities, must be equal to the work done by the friction force, along the distance d up the incline:

$W_{nc} = -\mu k*N*d$

The normal force, always normal to the surface, must be equal and opposite to the component of the weight normal to the incline:

$N = m*g*cos \theta = m*g*cos (24 deg) = \\ \\ 22.5 kg*9.8m/s2*0.913 = 201.4 N$

Replacing in the equation for Wnc:

$W_{nc} = -\mu k*N*d = -0.45*201.4 N*d = -90.6 N*d$

We can return to the equation (1) and solve for d:

$-208.1 J + 89.7N*d = -90.6N*d\\\\ d = \frac{208.1}{180.3} =1.15 m$

3 0

3 years ago

In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca

svlad2 [7]

Answer:

a) He found the same value of q/m for different cathode materials.

b) y = $- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}$ , c) v = $\frac{E_o}{B_o}$

Explanation:

In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.

b) In the part of the plates the electrons are accelerated by the electric field,

F = ma

- e E = m a

a = - (e/m) E₀

the distance traveled is

X axis

x = v₀ t

the separation of the plates is x = d

t = vo / d

Y axis

y = v_{oy} t + ½ to t²

y = ½ a t²

y = $- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}$

c) In this case there is a magnetic field B₀ and the electrons have no deflection

F = - e E + e v x B

if there is no deviation F = 0

e E = e v B

v = $\frac{E_o}{B_o}$

5 0

3 years ago

For each data set, which aspect of the data is likely to cause evidence obtained from it to be inconsistent?

sergij07 [2.7K]

I think it’s data set ll

4 0

3 years ago

Based on the macroscopic views of the three substances, which is most permeable?

zaharov [31]

Gas, because it's particles are not very close to each other when compared to the other two.

5 0

3 years ago

The standard unit of brightness is called the candela. True False