All categories

3 years ago

The end that points southward is called

Physics

1 answer:

Zepler [3.9K]3 years ago

7 0

Answer:

Explanation: The end of the magnet that points towards the south is called the north-seeking end or the south pole of the magnet.

You might be interested in

Plz I need help........................

Kazeer [188]

Answer:

Echolocation

Explanation:

.....................................

4 0

3 years ago

Read 2 more answers

A type O star is likely to appear

Ludmilka [50]

A type O star is likely to appear blue in color.

The blue type-O stars are solely thirty to fifty times a lot of large than the sun. However, O stars burn 1,000,000 times brighter and it has a very short lifespan. O stars solely last a couple of million years before they die in spectacular star explosions.

7 0

3 years ago

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from

anygoal [31]

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

$V_{max}=203 mph=90.74 m/s$ taking into account $1 mile=1609.34 m$

And the car's average acceleration $a_{ave}$ is:

$a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}$

Since:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}$ (1)

Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$ :

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$ , with the following equation:

$V_{max}=\frac{d_{remain}}{t_{2}}$ (11)

Isolating $t_{2}$ :

$t_{2}=\frac{d_{remain}}{V_{max}}$ (12)

$t_{2}=\frac{15362.35 m}{90.74 m/s}$ (13)

$t_{2}=169.45 s$ (14)

With this time $t_{2}$ and the value of $t_{1}$ calculated in (4) we can finally calculate the total time $t_{TOTAL}$ :

$t_{TOTAL}=t_{1}+ t_{2}$ (15)

$t_{TOTAL}=101.95 s + 169.45 s$ (16)

$t_{TOTAL}=271.4 s s$

5 0

4 years ago

The weld current between pulses is know as ______

Brut [27]

Answer:

the pulse arc system

Explanation:

6 0

3 years ago

If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what wo

Elena L [17]

Answer: The question is incomplete or some details are missing. Here is the complete question ; (a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.55 m/s2 for 4.05 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? m/s

(b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2

a ) With what speed (in m/s) does the car then strike the tree? m/s = 4.3125m/s

b) then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2 = -5.696m/s2

Explanation:

The detailed steps and calculation is as shown in the attached file.

6 0

4 years ago