All categories

2 years ago

Question 8 (4 points)

Physics

2 answers:

Paraphin [41]2 years ago

5 0

Answer:

S_d = 5 + 3*t

S_a = 4.2*t

t = 5 / 1.2 = 4.17 s

17.514 m

Explanation:

Given:

- The initial position of Duane s_i = 5

- The speed with which Duane runs v_d = 3 m/s

- The speed with which Albert runs v_a = 4.2 m/s

Find:

a) Write an expression for Duane's position in terms of t and vduane.

b. Write an expression for Albert's position in terms of t and valbert.

c. At what time does Albert catch up to Duane?

d. How far from Albert's starting point are they when they meet?

Solution:

- The position of Duane from start is:

S_d = s_i + v_d*t

S_d = 5 + 3*t

- The position of Albert from start is:

S_a = s_i + v_d*t

S_a = 0 + 4.2*t = 4.2*t

- When Albert catches up-to Duane their positions are equal.

S_a = S_d

5 + 3*t = 4.2*t

1.2*t = 5

t = 5 / 1.2 = 4.17 s

- The distance at which they meet is governed by time t calculated in part above:

S_a = 4.2*(4.17) = 17.514 m

katrin2010 [14]2 years ago

4 0

Answer:

The answer to your question is:

Explanation:

Data

Duane Albert

d = 5 m ; v = 3 m/s v = 4.2 m/s

a) b)

Duane's Albert's

d = 5 + (3)t d = 4.2t

d = 5 + 3t

c) 5 + 3t = 4.2t

4.2t - 3t = 5

1.2t = 5

t = 4.17 s

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

You might be interested in

An object is moving in the plane according to these parametric equations:

aniked [119]

A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π

b. vy = 4π cos (4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]

d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax

h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax

i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.

5 0

2 years ago

Read 2 more answers

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

2 years ago

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0

2 years ago

Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible

iren2701 [21]

Answer:

Option D

The air pressure inside the car is greater than the pressure outside.

Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

4 0

2 years ago

Mischka is an amateur astronomer. Using her telescope, she looks out past the Earth’s atmosphere and observes an object with a l

topjm [15]

A comet is long and also bright, which has a blue-ish color to it.

Best Regards, Mike

3 0

2 years ago

Read 2 more answers