All categories

2 years ago

Andy took a bus and then walked from his home to downtown. For the

Physics

1 answer:

natulia [17]2 years ago

8 0

Answer:

Av = 12.9 [km/h]

Explanation:

We must remember that the average speed is defined as the relationship between the space traveled over a certain time.

$Av = x/t$

where:

Av = average speed [km/h]

x = distance [km]

t = time [hr]

For when he was on the bus

The distance can be calculated by clearing the equation, for the term of the distance

$x_{1}= A_{v}*t\\x_{1}= 15*1.6\\x_{1}=24[km]$

For when Andy was walking

$x_{2}=A_{v}*t\\x_{2}=4.5*0.4\\x_{2}=1.8 [km]$

Now the total time is the sum of 1.6 + 0.4 = 2 [hr], and the total distance is the sum of the distances calculated above.

$x = 24 + 1.8\\x = 25.8 [km]$

The average speed is calculated using the ratio of space to time.

$Av = 25.8/2\\Av = 12.9 [km/h]$

You might be interested in

A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en

nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy $E_{n}$ of a particle of mass m in the nth energy state of an infinite square well potential with width L is:

$E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}$

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

$E_{1}=\frac{h^{2}}{8mL^{2}}$

$E_{2}=\frac{h^{2}}{2mL^{2}}$

So we can rewrite the energy in the ground state as:

$E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})$

$E_{1}=\frac{1}{4} E_{2}$

$E_{1}=\frac{1}{4} ( 6.0\ eV)$

Finally

$E_{1}=1.5\ eV$

6 0

2 years ago

carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr

denis23 [38]

The half-life equation $m=m_{0} (\frac{1}{2})^n$ in which n is equal to the number of half-lives that have passed can be altered to solve for n.

$n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}$

$\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2$

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

2 * 5700 = 11400 yr

3 0

2 years ago

Two vectors A and B are directed such that there is an angle θ between them. show answer Incorrect Answer Which of the following

Troyanec [42]

Answer:

Scalar product is between ║A║ ║ B║ and -║A║ ║ B║

Explanation:

Dot product between vec A and vec B is

A.B = ║A║ ║ B║ cos θ

Here, both ║A║ and ║B║ are positive and value of cos θ depends upon θ and lies between 1 and -1

So, Scalar product is between ║A║ ║ B║ and -║A║ ║ B║

8 0

2 years ago

A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

6 0

2 years ago

A future use of space stations may be to provide hospitals for severely burned persons. it is very painful for a badly burned pe

natta225 [31]

1.5 minutes per rotation. The formula for centripetal force is A = v^2/r where A = acceleration v = velocity r = radius So let's substitute the known values and solve for v. So F = v^2/r 0.98 m/s^2 = v^2/200 m 196 m^2/s^2 = v^2 14 m/s = v So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin. Its circumference is 2*pi*r, so C = 2 * 3.14159 * 200 m C = 1256.636 m And we need a velocity of 14 m/s, so 1256.636 m / 14 m/s = 89.75971429 s Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.

5 0

2 years ago