1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Leto [7]
3 years ago
7

Jason is taking a physical exam and has to do push ups for sixty seconds. He does 30 pushups in the sixty seconds. Each push up

takes two seconds to complete. Two seconds represents...
Physics
1 answer:
Alexandra [31]3 years ago
8 0

Answer: 2 seconds is the unit rate.

Explanation:

We know that Jason does 30 pushups in 60 seconds at a constant rate, and we know that each push up takes 2 seconds.

then, we can write this as 30 pushups/60 seconds = (1/2) pushups per second.

Here, two seconds represents the time needed to do one pushup, is the unit rate (this means that we need 2 seconds to have a unit "one pushup")

You might be interested in
All circuits include
valina [46]

Answer:

a battery, wires, and a switch.

Explanation:

All circuits include?

4 0
2 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
What is the mass of a liquid having a density of 1.50 g/ml and a volume of 3.5 liters?
PIT_PIT [208]

We know the formula for density = Mass/ volume

So     Mass, M = Volume * Density

         Volume = 3.5 L= 0.0035m^3

         Density = 1.50 g/ml  = 1500 kg/m^3

         Mass, M =  0.0035*1500 = 5.25 kg

So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.

6 0
4 years ago
A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg =  0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0
3 years ago
Calculate the force experienced by a positive test charge of 8.1 × 10-9 coulombs if the electric field strength is 9.4 × 107 new
Arte-miy333 [17]
The electric force exerted by an electric field of intensity E on a charge q is equal to the product between E and q, so:
F=qE=(8.1 \cdot 10^{-9}C)(9.4 \cdot 10^7 N/C)=0.76 N
5 0
3 years ago
Read 2 more answers
Other questions:
  • Write a short story about a time when you had to overcome the force of gravity to get something done
    14·1 answer
  • Which of the following is fact-based science rather than part of a personal belief system?
    8·1 answer
  • What is the significance of the nose end marking on a rocket or missile?
    7·1 answer
  • You are standing outside during the daytime. How can you see a blade of grass?
    8·2 answers
  • What is the critical angle of a light beam passed from a medium (n=2) to a medium (n=1.2)?
    14·1 answer
  • The saturated adiabatic lapse rate is a lesser lapse rate than the dry adiabatic lapse rate. this is because ________. the satur
    15·1 answer
  • As the speed/velocity of an object increases what happens to the kinetic energy of the object
    5·2 answers
  • What happens to ar object in free fall?​
    9·1 answer
  • A child has fallen while riding a bike. When should a bystander call 911 ?
    6·2 answers
  • For a constant voltage, how is the resistance related to the current?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!