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2 years ago

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Julian rides his bike uphill for 45 minutes, then turns around and rides back downhill. It takes him 15 minutes to get back to w

here he started. His uphill speed is 3 miles per hour slower than his downhill speed. Find Julian’s uphill and downhill speed.

1 answer:

Akimi4 [234]2 years ago

5 0

Answer:

1.5mi/hr, 4.5mi/hr

Explanation:

Given:

45 min = 0.75h, 15 min = 0.25h

(1) 0.75v₁ = 0.25v₂

(2) v₁ = v₂ - 3

Solve for v₁ and v₂:

0.75(v₂ - 3) = 0.25v₂ = 0.75v₂ - 2.25 = 0.25v₂

0.5v₂ = 2.25

v₂ = 4.5

v₁ = 1.5

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A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s

lbvjy [14]

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

4 0

2 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

2 years ago

A crane uses a block and tackle to lift a 2200N flagstone to a height of 25m

Cloud [144]

Remember the headline: ENERGY IS NEVER CREATED OR DESTROYED

The amount of energy before and after are always equal. All we ever do with energy is move it around from one place to another.

a). A crane can't create energy. Lifting the same rock in 20 different ways always takes the <u><em>same amount of work</em></u>. It doesn't matter whether one person picks the rock straight up, or 50 people get around it and lift it, or roll it up a ramp, or lift it with 16 pulleys and a mile of rope, or use a giant steam crane.

You want to lift a 2200N weight up 25m, you're going to have to supply

(2200N) x (25m) = <em>55,000 Joules</em> of work.

c). YOU put out 55,000 Joules of energy. It had to GO someplace. Where is it now ? ===> It's the potential energy the rock has now, from being 25m higher than it was before. That <em>55,000 Joules</em> is NOW the potential energy of the rock.

No energy was created or destroyed. It just got moved around.

55,000 Joules of energy began as nuclear energy in the core of the sun. Solar radiation carried it to the Earth. Plants absorbed it, and stored it as chemical energy. You ... or a cow that you ate later ... ate the plants and took the chemical energy. One way or the other, the chemical energy got stored in your blood and fat. When you needed to put it out somewhere, you moved it into your muscles, and they converted it into mechanical energy. Then you used the mechanical energy to exert forces. Today, you used the original 55,000 joules to lift the flagstone, and NOW that energy is in the flagstone, 25 meters up off the ground !

6 0

2 years ago

You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force

Tpy6a [65]

Answer:

Mass of Little Sister = 44.17 kg

Explanation:

From Newton's second law of motion, the magnitude of force applied on the sled is given by the following formula:

F = ma

where,

F = Force Applied = 120 N

a = Acceleration = 2.3 m/s²

m = Mass of Sled + Mass of Little Sister = 8 kg + Mass of Little Sister

Therefore,

120 N = (2.3 m/s²)(8 kg + Mass of Little Sister)

(120 N)/(2.3 m/s²) = 8 kg + Mass of Little Sister

Mass of Little Sister = 52.17 kg - 8 kg

<u>Mass of Little Sister = 44.17 kg</u>

4 0

2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

3 years ago