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zubka84 [21]
3 years ago
13

I was driving 150 km/hr for 20 km, then had to pull over for a quarter of an hour to replace my tire, and then drove the last 3.

5 hr at 200 km/hr. What was my final average velocity?
Physics
1 answer:
Likurg_2 [28]3 years ago
4 0

Answer:

ertyu

Explanation:

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A cat is running at 24 m/s. It then accelerates at 7 m/s2. How long will it take the cat to reach a speed of 49 m/s?
kozerog [31]

Answer:

t should be 3.57 second

Explanation:

Formula used is v = u+at

In which v is final velocity, u is initial velocity, a is acceleration and t is time.

Substitute each of the info given into the formula and calculate.

49 = 24 + (7)t

t = 3.57s

8 0
3 years ago
Which would exert the most gravity on a spacecraft that is passing close to Jupiter?
klio [65]
D. Jupiter has the highest amount of gravity in our solar system
7 0
2 years ago
A block of metal is 3 cm on each side. It has a mass of 283.5 grams. What is the density of the metal.
Luden [163]
Density =  \frac{Mass}{Volume}
\\\rho=\frac{m}{V}
\\\\\rho=\frac{283.5}{3^3}
\\\\\rho=\frac{283.5}{27}
\\\\\rho = 10.5  g/cm^3

5 0
3 years ago
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between th
Sauron [17]

Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

<u>V = 0.45 Volts</u>

4 0
3 years ago
A 50 kg crate is pulled across the ice by a rope that is angled at 30 degrees above the horizontal. If the tension in the rope i
denpristay [2]
Let:
Vx = the pulling component of force
Vy = the lifting component of force

Vy:
Sin(n°) = Vy/hypotenuse
hypotenuse * Sin(n°) = Vy
100N*sin(30°) = Vy
50N = Vy

Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
about 86.6N = Vx
4 0
3 years ago
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