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3 years ago

Question 7 of 10 What is cos(22")? O A. 0.93 B. 0.22 C. 0.37 O D. 0.40

Physics

2 answers:

8_murik_8 [283]3 years ago

8 0

Answer:

A. 0.93

Explanation:

Goshia [24]3 years ago

3 0

Answer:

Explanation:

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What is the primary way that a metamorphic rock form

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Metamorphic rocks form from the alteration of other rocks through pressure and temperature induced changes in the minerals.

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3 years ago

A 70.0-kilogram man is walking at a speed if 2.0 m/s. What is the kinetic energy?

Paraphin [41]

Answer:

140 J

Explanation:

From the the question, the mass of the man =70.0 kg and the speed at which the man is walking =2.0 m/s.

$K.E = \frac{1}{2}m {v}^{2}$

where K.E = the kinetic energy, m=mass and v= speed.

By substitution,

$K.E = \frac{1}{2} \times 70 \times {2}^{2}$

$\implies \: K.E = \frac{280}{2}$

$\implies K.E =140$

Hence the kinetic energy of the man is 140J

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3 years ago

Can you make sound you can see?

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3 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ (we use this approximation instead of the original $d^2+a^2\approx a^2$ since $z$ , our assumption still holds )

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

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4 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

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3 years ago