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myrzilka [38]
3 years ago
15

A car advertisement states that a certain car that Emily Miller love can accelerate from rest to 70km/h in 7 seconds. Fine the c

ar's average accelerate
Physics
1 answer:
alex41 [277]3 years ago
5 0

Average acceleration  =  (change in speed)  /  (time for the change)

                                       =           (70 km/hour)  /  (7 seconds)

                                       =               10  km/hr-sec .

That's a valid unit of acceleration ...  [ length / time² ] ...
but it's so clunky that we can't relate to it.  We usually
want to see acceleration in  m/s² .  So I'll convert this
answer to the more conventional unit.

                 (10 km/hr-sec) · (1,000 m / 1 km) · (1 hr / 3,600 sec)

             =  (10 · 1,000 · 1) / (1 · 3,600)  m/s²

             =          2.78  m/s²        ( about  0.28 G )
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What happens to energy and matter when a wave moves through space?
vivado [14]

Answer:

energy can move from one location to another, the particles of matter in the medium return to their fixed position. A wave transports its energy without transporting matter.

4 0
3 years ago
A bicycle of mass m requires 50 J of work to move from rest to a final speed v. If the same amount of work is performed during t
insens350 [35]

Answer:

The final speed of the second bicycle is (v·√2)/2

Explanation:

The mass of the given bicycle = m

The amount of work required to move the bicycle from rest to speed v = 50 J

The final speed of the first bicycle = v

The mass of the second bicycle = 2m

Therefore, from conservation of energy, we have;

Work required by the first bicycle = Kinetic energy gained by the bicycle

The kinetic energy = 1/2·m·v²

∴ Energy required by the first bicycle = 50 J = 1/2·m·v²

Given that the same amount of work is performed on the second bicycle, we have;

Work performed on the second bicycle = 50 J = kinetic energy of second bicycle = 1/2·(2·m)·v₂²

Also, given that 50 J = 1/2·m·v², we have;

Work performed on the second bicycle = 50 J = 1/2·m·v²= 1/2·(2·m)·v₂²

1/2·m·v²= 1/2·(2·m)·v₂²

m·v² = 2·m·v₂²

v² = 2·v₂²

v₂ = √(v²/2) = v/√2 = (v·√2)/2

v₂ = (v·√2)/2

The final speed of the second bicycle = v₂ = (v·√2)/2.

3 0
3 years ago
What will happen if an object or liquid with a density less than 1.0 g/cm3 is placed into water; do the same for an object or li
Anna71 [15]
If the object has a lower density than water, it will float.

If the object has a higher density than water, it will sink.
3 0
3 years ago
A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g
Alexus [3.1K]

Answer:

\Delta v= 4.66\frac{m}{s}

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system  the total momentum is constant if no external forces act in the system. The formula is:</em>

m_1v_1+m_2v_2=m_1u_1+m_2u_2

<em>Where:</em>

m_1: Mass of the first object.

m_2: Mass of the second object.

v_1: Initial velocity of the first object.

v_2: Initial velocity of the second object.

u_1: Final velocity of the first object.

u_2: Final velocity of the second object.

In <u>this problem</u> we have:

m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}

u_1=0\frac{m}{s}

Observation: v_1_2: Is because the system has the same initial velocity.

First we have to find u_2,

m_1v_1+m_2v_2=m_1u_1+m_2u_2

We can rewrite it as:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2

Replacing with the data:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}

Then, the resulting change in the cart speed is:

\Delta v= 4.66\frac{m}{s}

5 0
3 years ago
Determine the number of bonding electrons and the number of nonbonding electrons in the structure of of2.
Gnoma [55]
OF2 - 
<span>O has 6 electrons in outer shell and F has 7 in its outer shell </span>
<span>Therefore, you have to account for 20 electrons total in the </span>
<span>structure (7+7+6 = 20) </span>
<span>therefore draw it linear first. F ---- O-----F </span>
<span>The two bonds take care of 4 electrons now you have to add another 16. </span>
<span>Therefore 3 lone pairs on each F and 2 lone pair on O. </span>
<span>If you check for formal charges, all the atoms are neutral </span>
<span>F will have 3 lone pairs + 1 bond = 7 electrons (bond = 1/2 electron for formal charge distribution) therefore both the F's are neutral </span>
<span>Now look at the O: it should have 6.. it has two lone pair and 2 bonds = 4 electrons and 2 bonds = 1 electron each = 2 electrons from bonds = 6 total electrons for formal charge which is exactly the # it should have. There is no need for any double bond in this as there are no charges to be separated. </span>
<span>Now if u look at the # of domains around O you will see if you include the lone pairs it has a sp3 hybridization (4 domains) therefore a tetrahedron which has 2 lone pairs and 2 bonds.. since there are two lone pairs, the lone pair/bond pair repulsion is so high it is going to repel the two Fluorines and form a bent structure, looks a lot like H2O. </span>
4 0
3 years ago
Read 2 more answers
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