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2 years ago

Apply Newton’s second law to calculate the frictional force needed to hold a 15 g pen in your fingers and keep it from falling

Physics

1 answer:

Roman55 [17]2 years ago

3 0

Answer:

0.1962N.

Explanation:

Weight is a force, so we use newtons second law which states F=ma. So the weight of a watermelon on earth is 2kg x 9.81m/s², which is 19.62N (newtons). On the moon it weighs 2 x 1.63 which is 3.26N.

The weight of the pen is 0.02 x 9.81, which is 0.1962N. If the pen is not to fall the frictional force needs to be equal to this force,

so the answer is 0.1962N.

(Hope this helps can I pls have brainlist (crown)☺️)

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1) Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Bi

kifflom [539]

1) 13.7 m

The motion of the rock is a free fall, with constant acceleration g=9.8 m/s^2 towards the ground, so the total distance it covers is given by the SUVAT equation:

$S=\frac{1}{2}gt^2$

where S is the height of the house, and t is the time the rock takes to reach the ground. Substituting t=1.67 s, we find:

$S=\frac{1}{2}(9.8 m/s^2)(1.67 s)^2=13.7 m$

2) 105.5 m

The motion of the stuffed chicken is a projectile motion, with a uniform horizontal motion (with constant velocity of v=36.0 m/s) and a vertical accelerated motion (with constant acceleration of g=9.8 m/s^2).

First of all, we can find the total time of the ball by considering the vertical motion only. We know the vertical distance covered, S=42.2 m, so the time of the fall is

$S=\frac{1}{2}gt^2\\t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(42.2 m)}{9.8 m/s^2}}=2.93 s$

And now we can consider the horizontal motion to find the horizontal distance covered by the stuffed chicken:

$d=vt=(36.0 m/s)(2.93 s)=105.5 m$

3) 49.4 m

Again, the motion of the ball is a projectile motion, with a horizontal motion and a vertical motion.

The range of a projectile launched from the ground can be found by using the formula:

$d=\frac{v^2}{g} sin 2 \theta$

where, in this case:

v = 22.0 m/s is the initial velocity

$\theta=45^{\circ}$

Substituting into the formula, we find

$d=\frac{(22.0 m/s)^2}{9.8 m/s^2}(sin (2\cdot 45^{\circ}))=49.4 m$

4) 9.6 m/s^2

The frictional force acting on the monkey is given by:

$F_f = \mu mg=(0.16)(31.0 kg)(9.8 m/s^2)=48.6 N$

where $\mu$ is the coefficient of friction and m is the mass of the monkey.

We have two forces acting on the monkey: the push of F=345 N and the frictional force acting in the opposite direction. According to Newton's second law, the net force will be equal to the product between the monkey's mass and its acceleration, so we can find the acceleration:

$F-F_f=ma\\a=\frac{F-F_f}{m}=\frac{345 N-48.6 N}{31.0 kg}=9.6 m/s^2$

5) 462.3 N

The horizontal component of the pushing force is:

$F_x = F cos \theta = (648 N)(cos 25^{\circ})=587.3 N$

The frictional force, acting in the opposite direction, is

$F_f = \mu mg=(0.17)(75.0 kg)(9.8 m/s^2)=125.0 N$

where $\mu$ is the coefficient of friction and m is the mass of the box.

The net force on the box is therefore given by the net force on the horizontal direction:

$F_{net}=F_x -F_f=587.3 N -125.0 N=462.3 N$

6) 89.5 N

First of all we need to calculate the total weight of the table and the items above it.

The weight of the table is:

$W=mg=(25.0 kg)(9.8 m/s^2)=245 N$

So the total weight of the table and the items is

$W=245 N+63 N+12 N+44 N+24 N+9N+10N=407 N$

The force needed to get the table moving must be at least equal to the frictional force, which is equal to the product between the coefficient of friction and the weight of the all stuff:

$F=F_f = \mu W=(0.22)(407 N)=89.5 N$

7 0

2 years ago

A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three

yulyashka [42]

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

Speed of sound,v=343 m/s

We have to find the first three overtones.

Tube is closed

The overtone of closed pipe is equal to odd number of fundamental frequency.

Therefore, the overtone of tuba

$f'=nf$

Where n=3,5,7,..

Substitute n=3

$f'=3\times 88.4=265.2Hz$

For second overtone

$f'=5\times 88.4=442Hz$

For third overtone

$f'=7\times 88.4=618.8Hz$

4 0

3 years ago

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone

Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, $a_c$ , is given as follows;

$a_c = \dfrac{v^2}{r}$

Therefore, the centripetal acceleration of the stone found as follows;

$a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2$

The centripetal acceleration of the stone, $a_c$ = 5 m/s².

5 0

3 years ago

Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em

mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

E = h f

c= λ f

E = h c / λ

λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

#_photon = P / E₀

#_photon = 100 / 19.89 10⁻²⁰

#_photon = 5 10²⁰ photons / s

6 0

3 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

3 years ago

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