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2 years ago

14

A book moving across a desk has a net force equaling 60 N, to the right.

1 answer:

Maru [420]2 years ago

5 0

Answer:

80

Explanation:

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car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car

Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0

3 years ago

510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr

Yuki888 [10]

Answer:

The terminal velocity is $v_t =17.5 \ m/s$

Explanation:

From the question we are told that

The mass of the squirrel is $m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg$

The surface area is $A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2$

The height of fall is h =4.8 m

The length of the prism is $l = 23.2 = 0.232 \ m$

The width of the prism is $w = 11.6 = 0.116 \ m$

The terminal velocity is mathematically represented as

$v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }$

Where $\rho$ is the density of a rectangular prism with a constant values of $\rho = 1.21 \ kg/m^3$

$C$ is the drag coefficient for a horizontal skydiver with a value = 1

A is the area of the prism the squirrel is assumed to be which is mathematically represented as

$A = 0.116 * 0.232$

$A = 0.026912 \ m^2$

substituting values

$v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }$

$v_t =17.5 \ m/s$

7 0

3 years ago

The mass of the skier including his equipment is 75 kg in the ski race, the total vertical change in height is 880m

Bogdan [553]

75 percent off of water and please water the light water and water water and then go back and please water pollution please 880m

8 0

2 years ago

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

2 years ago

What are the charges of the subatomic particles?

kotegsom [21]

A particle smaller than an atom or a <span>cluster of such particles </span>

3 0

2 years ago