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Natasha_Volkova [10]
3 years ago
12

A 1800-kg Jeep travels along a straight 500-m portion of highway (from A to B) at a constant speed of 10 m/s. At B, the Jeep enc

ounters an unbanked curve of radius 50 m. The Jeep follows the road from B to C traveling at a constant speed of 10 m/s while the direction of the Jeep changes from east to south. What is the magnitude of the frictional force between the tires and the road as the Jeep negotiates the curve from B to C?
Physics
1 answer:
vladimir1956 [14]3 years ago
4 0

Answer:

3600N

Explanation:

F=m\frac{v^2}{r}

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A car is moving with a speed of 22 m/s. The driver then brakes, and the car comes to a halt after 6.5 s. What is the distance co
Shalnov [3]
The answer would be 72 miles. Hope this helped.
3 0
3 years ago
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two
natta225 [31]

Answer:

-15 m/s

Explanation:

The computation of the velocity of the 4.0 kg fragment is shown below:

For this question, we use the correlation of the momentum along with horizontal x axis

Given that

Weight of stationary shell = 6 kg

Other two fragments each = 1.0 kg

Angle = 60

Speed = 60 m/s

Based on the above information, the velocity = v is

1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0

\frac{60}{2} + \frac{60}{2} - 4\ v = 0

v = \frac{60}{4}

= -15 m/s

4 0
3 years ago
Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an
vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0
3 years ago
2 questions, brainliest if correct. please only answer if you really know!
Iteru [2.4K]

Answer:

<h2>1. Friction is A. a force</h2>

<h2>2. An unbalanced force is B. When the object moves and accelerates</h2>
3 0
3 years ago
Read 2 more answers
a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her
Rama09 [41]
Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2
5 0
4 years ago
Read 2 more answers
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