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V125BC [204]
2 years ago
13

A candy-filled piñata is hung from a tree for Matthew's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Hay

den cracks it with a 0.54-kg stick moving at 4.8 m/s. The stick stops and the piñata undergoes a gentle swinging motion. Determine the swing speed of the piñata immediately after being cracked by the stick.
Physics
1 answer:
Margaret [11]2 years ago
8 0

Answer:

The speed of the piñata immediately after being cracked by the stick is v_{fp}=0.59\: m/s.

Explanation:

Using the conservation of linear momentum:

m_{s}v_{is}=m_{p}v_{fp} (1)

Here:

m(s) is the mass of the stick

m(p) is the mass of the piñata

v(is) is the initial velocity of the stick

v(fp) is the final velocity of the piñata        

So, we just need to solve the equation (1) to v(fp).

v_{fp}=\frac{m_{s}v_{is}}{m_{p}}

v_{fp}=\frac{0.54*4.8}{4.4}

v_{fp}=0.59\: m/s

I hope it helps you!

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irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0
2 years ago
The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume
timurjin [86]

Answer:

P = 1 x 10⁸ Pa

Explanation:

given,

radius = 2.0 ×10⁻¹⁰ m

Temperature

T = 300 K

Volume of gas molecule =

V = \dfrac{4}{3}\pi r^3

V = \dfrac{4}{3}\pi (2\times 10^{-10})^3

 V = 33.51 x 10⁻³⁰ m³

we know,

P  V = 1 . k T

k = 1.38  x 10⁻²³ J/K

P(33.51 x 10⁻³⁰) = 1 . (1.38  x 10⁻²³) x 300

P = 1.235 x 10⁸ Pa

for 1 significant figure

P = 1 x 10⁸ Pa

6 0
2 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

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k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
An 8 Ω resistor is connected to a 6 V battery and another resistor in series. If the current in the circuit is 0.5 A, what is th
Semmy [17]
8+R=V/I
8+R=6/0.5=12 ohm
then R=4 ohm
4 0
2 years ago
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