What is the weight of a 20 kg box on the Earth?<br> A) 2 N <br> B) 20 N<br> C) 200 N

A wood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above hor

Stells [14]

The solution for the problem is:

Constant speed means Fnet = 0.
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:

Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ)
mg (sin Θ) = µk (mg) (cos Θ)
µk = tan Θ
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

5 0

4 years ago

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

5 0

3 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

A puck of mass 0.5100.510kg is attached to the end of a cord 0.8270.827m long. The puck moves in a horizontal circle without fri

yanalaym [24]

Answer: 2.75 1/sec

Explanation:

The only external force (neglecting gravity) acting on the puck, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r (1)

Our unknown, is the frequency at which the puck can go around the circle, which is the inverse of the period Tp.

By definition, a period is the time needed by the puck to complete one entire circle.

By definition also , angular velocity is the rate of change of the angle advanced, so we can express this way:

ω = ∆θ / ∆t

The angle advanced during one period, is exactly (by angle definition) 2 π radians.

So, we can always write the angular velocity, ω, as follows:

ω = 2π / Tp = 2πf

Now, there is a relationship between linear and angular velocity, that can be found applying simply the definition of velocity and of an angle too, as follows:

v = ∆s / ∆t = r ∆θ/∆t = ω r

Replacing in (1), we have:

T = mω2 r2 / r = m ω2r (2)

We have just found that ω= 2πf, so, replacing in (2) :

T = m (2π)2 f2 r

Solving for f:

f = 1/2π√(T/mr) = 1/2π 17.28 1/sec = 2.75 1/sec

6 0

3 years ago

Please help!

andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0

3 years ago

What is the weight of a 20 kg box on the Earth? A) 2 N B) 20 N C) 200 N