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2 years ago

Type the correct answer in the box.

Physics

2 answers:

vfiekz [6]2 years ago

7 0

Answer:
64.4 degrees Fahrenheit

Hope this helps!

sleet_krkn [62]2 years ago

4 0

Answer:

Explanation:

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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

3 years ago

What is the frequency of an X-ray with wavelength 0.13 nm ? Assume that the wave travels in free space. Express your answer to t

dem82 [27]

Answer:

Frequency, $f=2.30\times 10^{18}\ Hz$

Explanation:

Given that,

The wavelength of the x-rays, $\lambda=0.13\ nm=0.13\times 10^{-9}\ m$

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

$c=f\lambda$

f is the frequency

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz$

So, the frequency of an x-ray is $2.30\times 10^{18}\ Hz$ . Hence, this is the required solution.

5 0

2 years ago

A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

larisa86 [58]

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

8 0

3 years ago

Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati

jek_recluse [69]

Answer:

$a=4\ m/s^2$

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2$

So, the magnitude of centripetal acceleration is $4\ m/s^2$ .

5 0

2 years ago

A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch

ololo11 [35]

Answer:

$v_{1f}$ = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

p₀ = (m₁ + m₂) v₀

After the explosion

pf = m₁ $v_{1f}$ + m₂ $v_{2f}$

p₀ = [texpv_{f}[/tex]

(m₁ + m₂) v₀ = m₁ $v_{1f}$ + m₂ $v_{2f}$

Let's calculate the final speed (v1f) of the first stage

$v_{1f}$ = ((m₁ + m₂) v₀ - m₂ $v_{2f}$ ) / m₁

$v_{1f}$ = ((2100 +1160) 4300 - 1160 5940) / 2100

$v_{1f}$ = (14,018 10 6 - 6,890 106) / 2100

$v_{1f}$ = 7,128 106/2100

$v_{1f}$ = +3,394 103 m / s

come the same direction of the final stage, but more slowly

4 0

3 years ago