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3 years ago

14

A person runs 400 meters in a straight line. Determine their distance and displacement. Is the distance equal to the displacemen

t for a person that runs 2.5 times around a circular track?

1 answer:

Levart [38]3 years ago

4 0

Answer:

their distance and displacment are both 400 meters. and No distance and displacement are not equal for a person who runs 2.5 times a circular track.

Explanation:

if you run in a ircle you are coming back to the starting poing ad remember distane is the total length travlled while displacement is the final- initial length

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A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

3 years ago

What is buoyant force?

Nostrana [21]

This is the upthrust on an object which is placed inside a fluid

This force act upwards and always push upwards

so the correct answer is given as

D. A force within a fluid that pushes upward

this force is always due to pressure difference at two levels of

at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy

3 0

3 years ago

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

So

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is $\sqrt{3^2+(-5)^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$

3 0

3 years ago

The physical structure of the earth’s rock is changed by _____.

PSYCHO15rus [73]

It could be stress or strain

4 0

4 years ago

Read 2 more answers

Desperate for help! please!!!!!!!!

lys-0071 [83]

$0.4029 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$

$\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}$

8 0

3 years ago