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1 year ago

A 70.0Kg cyclist Rides a 15.0kg

Physics

1 answer:

Inessa [10]1 year ago

4 0

Answer:

$1275 kg*m/s$

Explanation:

We'll use the momentum equation:

$p=mv$

where:

p = momentum

m = mass

v = velocity

Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

$70.0 kg + 15.0kg=85.0 kg$

Given that, we can now substitute our given values into the momentum equation:

$p=mv\\p=85.0kg*15.0m/s$

Our final answer is:

$p= 1275 kg*m/s$

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A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie

cluponka [151]

Answer:

a) v₃ = 19.54 km, b) 70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

cos 45 = x₁ / V₁

sin 45 = y₁ / V₁

x₁ = v₁ cos 45

y₁ = v₁ sin 45

x₁ = 26 cos 45

y₁ = 26 sin 45

x₁ = 18.38 km

y₁ = 18.38 km

Vector 2 moves 45 km north

y₂ = 45 km

Unknown 3 vector

x3 =?

y3 =?

Vector Resulting 70 km north of the starting point

R_y = 70 km

we make the sum on each axis

X axis

Rₓ = x₁ + x₃

x₃ = Rₓ -x₁

x₃ = 0 - 18.38

x₃ = -18.38 km

Y Axis

R_y = y₁ + y₂ + y₃

y₃ = R_y - y₁ -y₂

y₃ = 70 -18.38 - 45

y₃ = 6.62 km

the vector of the third leg of the journey is

v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

v₃ = √ (18.38² + 6.62²)

v₃ = 19.54 km

to find the angle let's use trigonometry

tan θ = y₃ / x₃

θ = tan⁻¹ (y₃ / x₃)

θ = tan⁻¹ (6.62 / (- 18.38))

θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

θ’= 180 -19.8

θ’= 160.19º

I mean the address is

θ’’ = 90-19.8

θ = 70.2º

70.2º north-west

3 0

2 years ago

If you travel at 3 m/s for 12 seconds, how far did you travel?

ella [17]

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

4 0

2 years ago

Read 2 more answers

An extension cord is used with an electric weed trimmer that has a resistance of 17.9 Ω. The extension cord is made of copper (r

Naddika [18.5K]

Answer:

(a) $R_{c}=0.87ohms$

(b) $V_{T}=114.44V$

Explanation:

Part (a)

The total length of copper cord L=86.3 m

The cross sectional area A=1.71×10⁻⁶m²

The resistivity of copper p=1.72×10⁻⁸Ω

Thus the resistance of extension cord is

$R_{c}=p\frac{L}{A}\\R_{c}=(1.72*10^{-8} )\frac{86.3}{1.71*10^{-6}}\\R_{c}=0.87ohms$

Part (b)

The resistance of trimmer Rt=17.9 ohms

When voltage of 120V is applied then the current I is passing through series circuit is

$I=\frac{120V}{R_{c} +R_{T} }\\I=\frac{120V}{0.87 +17.9 } \\I=6.4A$

Thus the voltage across the trimmer is:

$V_{T}=IR_{T}\\V_{T}=(6.4)*(17.9)\\V_{T}=114.44V$

8 0

2 years ago

Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?

Ainat [17]

PV = 400 x 0.08 = 32 J

Hope this helps

7 0

2 years ago

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

<em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

$T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\ \times \ years}{3600 \ s \ \times\ 24\ h\ \times \ 365.25 \ days} \\\\T = 1.397 \times 10^{10} \ years\\\\T = 13.97 \ billion \ years$

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0

2 years ago