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Vaselesa [24]
3 years ago
15

HELP ME PLEASE

Physics
1 answer:
olga55 [171]3 years ago
4 0

1) False

2) True

3) The distance is 17 m

4) The displacement is 2 m south

5) She did not give the direction as either left, or negative.

6) The average speed is 28 mph

7) The average velocity is 12 mph north

8) The average speed is 0.27 m/s

9) False

10) False

Explanation:

1)

Speed is a scalar quantity which tells how fast an object is moving regardless of its direction, and it is calculated as:

speed=\frac{d}{t}

where d is the distance covered by the object and t is the time taken. Being a scalar quantity, speed consists only of a value and its units, so no direction needs to be specified.

2)

Velocity is a vector quantity, defined as

velocity = \frac{d}{t}

where d is the displacement of the object (a vector connecting the initial position to the final position of motion) and t is the time taken. Being a vector, velocity has both a magnitude and a direction (the same direction as the displacement), so direction here should also be specified.

3)

The distance travelled by the object is just the total length of the path taken, regardless of the direction of each part of the motion.

Here the object moves:

10 meters to the right

7 meters to the left

So, the distance travelled is

d = 10 + 7 = 17 m

4)

The displacement is a vector connecting the initial position to the final position of motion, so we have to compare the starting position with the final position.

Taking x = 0 as initial position, and north as positive direction:

- The object moves 5 m north first (+5)

- The object moves 7 m south (-7)

So, the displacement is

d = +5 + (-7) = -2 m

which means 2 meters south.

5)

As we said previously, displacement is a vector connecting the initial position to the final position of motion. Being a vector, it must have:

- A magnitude (the shortest distance between the initial and final position, in a straight line)

- A direction

Here Jenny reported only the magnitude (3 meters), but not the direction, so she forgot to include the direction of the displacement (which is to the left).

6)

The average speed is given by

speed=\frac{d}{t}

where d is the distance and t is the time taken.

The distance is just the total length covered, so:

d = 5 + 2 = 7 miles

The time taken is

t = 1/4 h = 0.25 h

So, the average speed is

speed=\frac{7}{0.25}=28 mph

7)

The average velocity is given by

velocity=\frac{d}{t}

where d is the displacement and t is the time taken.

The displacement is, taking north as positive direction:

d = +5 + (-2) = 3 miles (north)

The time taken is

t = 1/4 h = 0.25 h

So, the average velocity is

velocity=\frac{3}{0.25}=12 mph (north)

8)

We can calculate the average speed by adding the single measurements and dividing by the number of trials done:

speed_{avg}=\frac{s_1+s_2+s_3}{3}

where in this case, N = 3. For this experiment we have:

s_1 = \frac{0.75 m}{2.5s}=0.30 m/s\\s_2 = \frac{0.75 m}{2.75 s}=0.27 m/s\\s_3=\frac{0.75 m}{2.98 s}=0.25 s

So the average is

speed_{avg}=\frac{0.30+0.27+0.25}{3}=0.27 m/s

9)

Data are said to be:

- Accurate, when the average value of the measurements is close to the actual value

- Precise, when the spread of the measurements done in the different trials is small

Therefore, the first part of the sentence "Data that is accurate, is data that is really close to the actual value" is correct, and the second part "data that is precise is data that is repeated over and over again" is not correct, since we may have several measurements but their spread may be large.

10)

As we said in part 9):

- Accuracy refers to how close the measured value is to the actual value

- Precision refers to the spread (or the uncertainty) on the measured value: the smaller it is, the better the precision

For instance, let's assume that the actual value of a certain variable is 3.0. If we get the following set of data:

2.4, 2.5, 2.4, 2.3

it is precise (the spread is small) but not accurate (since the average, 2.4, is far from the actual value)

while the following set:

3.1, 3.6, 2.4, 3.0

is accurate (the average is around 3.0, so close to the actual value), but not precise (the spread is very large).

Learn more about speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

brainly.com/question/5248528

#LearnwithBrainly

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A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the p
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Answer:

Explanation:

Given that,

Charge q=-5.90nC

Magnetic field B= -1.28T k

And the magnetic force

F =−( 3.70×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

The force in a magnetic field is given as

F= q(v×B)

−( 3.70×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

x=-100.64m/s

a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

3.724×10^-5 - 3.724×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cosx

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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Answer:

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t=\frac{d}{v}=\frac{60 m}{19.97 s}=3.0 s

8 0
3 years ago
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