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Mila [183]
3 years ago
6

What is the speed of a car that travels 14 meters in 2 seconds?

Physics
1 answer:
mars1129 [50]3 years ago
8 0

Distance traveled by car =14m

Total time taken =2sec

Speed =distance / time

=14m/2sec

speed=7 m/sec


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A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe
Bogdan [553]

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

4 0
3 years ago
The wave speed of sound waves traveling through four different media was measured at a constant temperature. Results are summari
Kitty [74]

Answer:b

Explanation:

6 0
3 years ago
A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (ve
Lemur [1.5K]

The Force on the left hand pole, F' = 0.167N

<h3>What is the force on the left hand pole?</h3>

Force is an agent which produces a change in the motion or state of an object.

Force is a vector quantity.

The general force is calculated as follows:

F = mg/sinθ

m = 17.1 g = 0.0171 kg

g = 9.81 m/s²

θ = 45°

F = 0.0171 * 9.81/sin45

F = 0.237 N

Force on the left hand pole, F' = Fcosθ

F' = 0.237 * cos 45

F' = 0.167N

In conclusion, the force on the left hand pole is the horizontal component of force.

Learn more about force at: brainly.com/question/141439

#SPJ1

5 0
2 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
Charlotte is driving at 66.5 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 3
Alborosie

Answer:

the distance traveled by Charlotte in feet is 338.44 ft

Explanation:

Given;

speed of Charlotte, u = 66.5 mi/h

time of motion, t = 3.47 s

The distance traveled by Charlotte in feet is calculated as;

Distance = Speed \ \times \ time \\\\D = ut\\\\D = (\frac{66.5 \ mi}{h} \times  \frac{5280 \ ft}{1 \ mi} \times \frac{1 \ h}{3600 \ s} )(3.47 \ s)\\\\D = 338.44 \ ft

Therefore, the distance traveled by Charlotte in feet is 338.44 ft

7 0
3 years ago
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