PHYSICS COLLEGE

Impulse is the______of the force and time of contact

Answers

Answer 1

Answer:

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

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What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

A skater has rotational inertia 4.2 kg-m2 with his fists held to his chest and 5.7 kg?m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they�re essentially on his rotation axis, how fast will he be spinning? Express your answer using two significant figures. ?f=

Answers

Answer: 38.5rad/s

Explanation: The calculations can be viewed on the image attached below. Thanks

In a Venn diagram, the separate circles contain characteristics unique to each item being compared and the intersection contains characteristics that are common to both items being compared. Ernie is working on the Venn diagram below to compare the career pathways of Biotechnology Research and Development and Diagnostic Services.What else could Ernie put in the common section?

Collecting data and analyzing results
Designing and implementing systems
Maintaining and using diagnostic equipment
Designing and using laboratory equipment
Mark this and return

Answers

Another thing that Ernie put in the common section is collecting data and analyzing results.

What is a Venn diagram?

A Venn diagram is used to show a representation of data. The center of the Venn diagram is often used to indicate the data set that is the same.

Looking at the Venn diagram, another thing that Ernie put in the common section is collecting data and analyzing results.

Learn more about Venn diagrams:brainly.com/question/1605100

#SPJ2

I think its- Collecting data and analyzing results
Thats what I put.

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads share the 28 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 N . Assume that A has a greater charge. What is the charge qA and qB on the beads?

Answers

Answer:

Explanation:

Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = $(9*10^9* q*(28-q)*10^(-18))/((5*10^(-2))^2)$

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)