All categories

3 years ago

Please someone help me complete this.. I need it done before 9am!!!

Physics

2 answers:

Shalnov [3]3 years ago

7 0

Answer:

hi i hope it helps

Explanation:

so ,

1) bike

2) runner

3) Walker

CaHeK987 [17]3 years ago

7 0

1 bike 2 runner 3 walker

You might be interested in

Suppose the B string on a guitar is 24" long and vibrates with a frequency of about 247 Hz. You place your finger on the 5th fre

Rudiy27

Answer:

b. E (about 329 Hz)

Explanation:

Given data:

Initial length of the string l1= 24 in

initial frequency f1= 247 Hz

changed length l2= 18 in

Then we have to find the changed frequency f2= ?

We already now that

frequency f ∝ 1/length of the string l

therefore,

$\frac{f_1}{f_2} =\frac{l_1}{l_2}$

⇒ ${f_2}=\frac{l_1}{l_2}\times{f_1}$

⇒ ${f_2}=\frac{24}{18}\times{247}$

⇒ ${f_2}=329.33 Hz$

7 0

3 years ago

An elements atomic number is 79. How many protrons would an atom of this element have?

Andrews [41]

9 thats you electron number and what makes up 9 is 3

6 0

3 years ago

Which of the following statements about Boyle's law are correct? The temperature and pressure must remain constant for the law t

kvasek [131]

Answer: The temperature and the number of molecules must reamain constant for the law to apply, and as the pressure increases, the volumen decreases proportionally.

Boyle's law states that if the temperature, T, of a given mass of gas, remains constant, the Volume, V, of the gas is in inverse relation to the pressure, p; i.e.

pV = constant (for a given mass of gas, at constant T)

Then, if p increases, V decreases proportionally to keep the relation pV = constant.

8 0

3 years ago

Read 2 more answers

The elementary particle called a muon is unstable and decays in about 2.20μs2.20μs , as observed in its rest frame, into an elec

vredina [299]

Answer:

5.865 μs

Explanation:

t₀ = Time taken to decay a muon = 2.20 μs

c = Speed of Light in vacuum = 3×10⁸ m/s

v = Velocity of muon = 0.927 c

t = Lifetime observed

Time dilation

$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds$

∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs

3 0

3 years ago

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis o

mihalych1998 [28]

Answer:

θ₁ = 0.5 revolution

Explanation:

We will use the conservation of angular momentum as follows:

$L_1=L_2\\I_1\omega_1=I_2\omega_2$

where,

I₁ = initial moment of inertia = 18 kg.m²

I₂ = Final moment of inertia = 3.6 kg.m²

ω₁ = initial angular velocity = ?

ω₂ = Final Angular velocity = $\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}$ = 1.67 rev/s

Therefore,

$(18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1$

where,

θ₁ = revolutions if she had not tucked at all = ?

t₁ = time = 1.5 s

Therefore,

$\theta_1 = (0.333\ rev/s)(1.5\ s)\\$

<u>θ₁ = 0.5 revolution</u>

4 0

3 years ago