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AleksAgata [21]

2 years ago

11

A race car driver must average 200km/hr for four laps to qualify for a race. Because of engine trouble, the car averages only 17

0km/hr over the first two laps. What is the average speed must be maintained for the last two laps

1 answer:

vampirchik [111]2 years ago

5 0

The average speed would have to be 260 km/hr due to the driver originally going 30 km/hr too slow the first two laps

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A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

Where,

F = Force/Weight

A = Area

$\Delta L$ = Compression

$L_0$ = Original Length

According to the values given we have to

$\Upsilon_{steel} = 200*10^9Pa$

$\Delta L = 5.6*10^{-7}m$

$L_0 = 3.2m$

$r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2$

Replacing this values at our previous equation we have,

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

$200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}$

$F = 38272.5N$

Therefore the Weight of the object is 3.82kN

4 0

2 years ago

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

Vikki [24]

acceleration of rocket is given here as

$a_y = 2.60* t$

now we know that

$\frac{dv}{dt} = 2.60t$

now integrating both sides

$\int dv = \int 2.60t dt$

$v = 2.60\frac{t^2}{2}$

$v = 1.30 t^2$

here since its given that rocket will accelerate for t = 10 s

so here we have

$v = 1.30 * 10^2$

$v = 130 m/s$

so after t = 10 s the speed of rocket will be 130 m/s upwards

5 0

2 years ago

A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

2 years ago

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

L = W

Area

A = W W = W²

R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

Length L = D= 2W

Area A = W L

R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

R₂ / R₁ = 2W²/L

6 0

2 years ago

A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = $\frac{\pi d^{2} }{4}$

where d is the diameter of the wire

1.613 x 10^-7 = $\frac{3.142* d^{2} }{4}$

6.448 x 10^-7 = 3.142 x $d^{2}$

$d^{2}$ = $\sqrt{ 2.05*10^-7}$

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0

2 years ago