Please help on this one?

Vivian went on a bicycle trip through Germany with her family. One afternoon, she rode her bicycle along a long flat road at a c

natima [27]

d = distance traveled by her on her bicycle on a long flat road = 24 kilometer

t = time taken by her to travel distance "d" on her bicycle on a long flat road = 1.2 hours

v = average speed of vivian = ?

we know that average speed is given as

v = d/t

inserting the values in the above formula

v = 24 kilometer / 1.2 hour

v = 20 kilometer/hour

hence the correct choice is

C) 20 km/h

8 0

2 years ago

How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated

guapka [62]

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

5 0

2 years ago

A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from th

Ipatiy [6.2K]

Answer:

$N_{2}$ =20 turns

Explanation:

The given case is a step down transformer as we need to reduce 120 V to 6 V.

number of turns on primary coil N_{P}= 400

current delivered by secondary coil I_{S}= 500 mA

output voltage = 6 V (rms)

we know that

$I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}$

putting values we get

$I_{p}=\frac{6}{120\times 0.5}$

$I_{p}= 0.1 A$

to calculate number of turns in secondary

$\frac{N_{2}}{400} =\frac{6}{120}$

therefore, $N_{2}$ =20 turns

5 0

2 years ago

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

2 years ago

Determine whether the given value is a statistic or a parameter. a homeowner measured the voltage supplied to his home on 20 day

Gelneren [198K]

<span>For this example, the value presented would be considered a statistic. The value is a statistic as it represents a numerical measurement of a sample. If it were a parameter, it would need to represent a numerical measurement of a population.</span>

4 0

2 years ago