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makkiz [27]
3 years ago
12

What will occur when two have reach one another

Physics
1 answer:
grigory [225]3 years ago
5 0
Well, you would have to be more specific! Sorry I couldn't help you, but if you have the whole question and it is udnerstandable, please repost your question!

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A home uses ten 100-watt lightbulbs for five hours per day. Approximately how many kilowatt-hours of electrical energy are consu
timurjin [86]

The electrical energy consumed in one year by using the light bulbs is 8,760 kWh.

The given parameters:

  • Number of light bulbs = 10
  • Power consumed by each bulb = 100 W
  • Time of energy consumption, t = 1 year

<h3>What is electrical energy?</h3>

This is the electric power consumed or dissipated at a given period of time.

The  electrical energy consumed in one year by using the light bulbs is calculated as

E = Pt

E = (100 \times 10) \times (1 \ yr \times \frac{8760 \ hrs}{1 \ yr} )= 8,760,000 \ W att-hours\\\\&#10;E = \frac{8,760,000 }{1000} = 8,760 \ kWh

Thus, the electrical energy consumed in one year by using the light bulbs is 8,760 kWh.

Learn more about electrical energy here: brainly.com/question/60890

4 0
2 years ago
Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity
AfilCa [17]

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ4

5 0
3 years ago
An object is in simple harmonic motion. Find an equation for the motion given that the frequency is 3⁄π and at time t = 0, y = 0
Delicious77 [7]

Answer: y(t)= 1/π^2 sin(6*π^2*t)

Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:

y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.

The data are: ν= 3π; y(t=0)=0 and y'(0)=6.

Firstly we know that 2πν=ω then ω=6*π^2

Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0

Besides y'(t)=6*π^2*A*cos (6*π^2*t)

y'(0)=6=6*π^2*A*cos (6*π^2*0)

6=6*π^2*A then A= 1/π^2

Finally the equation is:

y(t)= 1/π^2 sin(6*π^2*t)

3 0
3 years ago
Read 2 more answers
A 1000-kg car traveling north at 15 m/s collides with a2000-kg truck traveling east at 10 m/s. The occupants, wearing seat belts
Aleksandr-060686 [28]

Answer:

8.33 m/s, 36.87° North of East

Explanation:

m_n = Mass of car = 1000 kg

v_n = Velocity of car = 15 m/s

m_e = Mass of truck = 2000 kg

v_e = Velocity of truck = 10 m/s

M = Combined mass = 1000+2000 = 3000 kg

Momentum

p_n=m_nv_n\\\Rightarrow p_n=1000\times 15\\\Rightarrow p_n=15000\ kgm/s

Momentum of car traveling East is 15000 kgm/s

p_e=m_ev_e\\\Rightarrow p_n=2000\times 10\\\Rightarrow p_n=20000\ kgm/s

Momentum of truck traveling North is 20000 kgm/s

Angle

\theta=tan^{-1}\frac{p_n}{p_e}\\\Rightarrow \theta=tan^{-1}\frac{15000}{20000}\\\Rightarrow \theta=36.87^{\circ}

As the two vehicles are vectors, the resultant velocity is

(Mv)^2=p_n^2+p_e^2\\\Rightarrow v=\sqrt{\frac{p_n^2+p_e^2}{M^2}}\\\Rightarrow v=\sqrt{\frac{15000^2+20000^2}{3000^2}}\\\Rightarrow v=8.33\ m/s

Velocity of the two vehicles when they are locked together is 8.33 m/s and direction is 36.87° North of East

5 0
3 years ago
The frequency of the fundamental of the guitar string is 320 Hz. At what speed c do waves move along that string?
viktelen [127]

Complete question is The frequency of the fundamental of the guitar string is 320 Hz. At what speed c do waves move along that string?  wavelength is 40 cm.

Answer:

128 m/s  

Explanation:

In case where fundamental frequency is given, the speed with waves travel along the string can be calculated using the following formula:

v = f (2L) where L is the length of the string (L = λ/2)

⇒v= f λ

f = 320 Hz (given)

λ = 40 cm = 0.40 m

Substitute the values:

⇒ v = 320 Hz × 0.40 m=  128 m/s

7 0
3 years ago
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