What is the speed of a bobsled whose distance-time graph indicates that it traveled 114m in 30s? m/s

A dog sits 2.6 m from the center of a merry- go-round. a) If the dog undergoes a 1.7 m/s^2 centripetal acceleration, what is the

alexgriva [62]

Answer:

a) v= 2.1 m/s

b) ω = 0.807 rad/s

Explanation

Conceptual analysis :

The dog and the merry-go- round describes a circular motion, then, the following formulas apply :

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

v = ω *r Formula (2)

Where:

$a_{c}$ : Centripetal acceleration(m/s²)

v: linear speed or tangential (m/s)

r : radius of the circle (m)

ω : angular speed ( rad/s)

Data

r= 2.6 m

$a_{c}$ = 1.7 m/s²

Problem develpment

a) We replace data in the formula 1 to calculate the dog's linear speed(v):

$a_{c} =\frac{v^{2} }{r}$

$1.7 =\frac{v^{2} }{2.6}$

$v^{2} =1.7*2.6 = 4.42$

$v=(\sqrt{4.42})\frac{m}{s}$

v= 2.1 m/s

b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).

v = ω *r

2.1 = ω *2.6

ω = 2.1/2.6

ω = 0.807 rad/s

6 0

2 years ago

An unnamed professor is walking to class when an office chair (travelling in the same direction) collides with them. The profess

ololo11 [35]

Answer: 1.96 m/s

Explanation:

Given

Mass of Professor $m_1=113\ kg$

Velocity of professor $u_1=1.56\ m/s$

mass of chair $m_2=10\ kg$

velocity of chair $u_2=6.5\ m/s$

Suppose after the collision, v is the common velocity

Conserving momentum

$\Rightarrow m_1u_1+m_2u_2=(m_1+m_2)v\\\\\Rightarrow v=\dfrac{113\times 1.56+10\times 6.5}{113+10}=\dfrac{241.28}{123}\\\\\Rightarrow v=1.96\ m/s$

5 0

2 years ago

What is the velocity of a wave with a wavelength of 9 meters and a period of 0.006

Ipatiy [6.2K]

Answer: 1,500m/s

Explanation:

Relationship existing between velocity of a wave (v), wavelength(¶) and frequency(f) is

v = f¶... (1)

Since Frequency (f) is the reciprocal of the period (T);

Frequency = 1/Period i.e F = 1/T... (2)

Substituting equation 2 into 1 we have;

v = 1/T × ¶

v = ¶/T

Given wavelength ¶ = 9m

Period T = 0.006s

v = 9/0.006

v = 1,500m/s

The velocity of the wave will be 1,500m/s

8 0

2 years ago

In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the

AleksandrR [38]

Answer:

The charge on the drop is

q = 1.741 x 10 ⁻²¹ C

Explanation:

Electric field due to plates

Ef = V/d

Ef = 2033 V / (2.0 * 10^-2 m )

Ef = 101650 V/m

So, we can write

Ef * q = m*g

q = m*g / E f

The mass can be equal using the density and the volume so:

m = ρ * v

The volume can be find as:

v = 2.298 x 10 ⁻ ¹⁶ m³

q = ρ * v * g / Ef

q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

The charge on the drop is

q = 1.741 x 10 ⁻²¹ C

5 0

2 years ago

It takes 20s for 3x10^6 electrons to flow through a wire ,what is the current?

Elina [12.6K]

Answer:

Explanation:

The variables we know and are given are:

time, t = 20s

Charge, Q = 3x1-^-6 electrons, which is just 3x10^-6C (C stands for Coulombs, which is the unit for Charge)

We need to find the current, I, and since we know Q and t we can substitute these values into the given equation:

I=Q/t (which if you look at what the RHS is saying, its Charge over time, or more literally means the amount of charge passing a point over a period of time)

If we substitute these values, we will get I as:

I = Q / t

I = 3x10^-6 / 20

I = 1.5x10^-7 A

Hope this helps!

6 0

2 years ago