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d1i1m1o1n [39]
2 years ago
14

A car is moving 18 m/s to the east. if it takes the car 5 seconds to reach a velocity of 19 m/s to the east, what is its acceler

ation?
A. 1 m/s2 east
B. 0.2 m/s2 west
C. 0.2 m/s2 east
D. 5 m/s2 east
Physics
2 answers:
almond37 [142]2 years ago
5 0
<span>C. 0.2 m/s2 east

Hope it helps!
</span>
kakasveta [241]2 years ago
4 0
<h2>Option C is the correct answer.</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 18 m/s to east

Final velocity, v = 19 m/s to east

Acceleration, a = ?

Time, t = 5 s

Substituting

             v = u + at

             19 = 18 + a x 5

              1 = 5a

               a = 0.2 m/s²

Since the velocity direction is towards east, direction of acceleration is also towards east.

So acceleration is 0.2 m/s² towards east.

Option C is the correct answer.

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an 11 kg tool box is floating stationary in an orbiting spacecraft when a 79 kg astronaut, initially at rest, gives the tool box
snow_tiger [21]

The astronaut final velocity is 0.06 m/s to the right

Explanation:

In absence of external forces, the total momentum of the box-astronaut system is conserved.

At the beginnig, the total momentum of the two is zero, since they are at rest:

p_i = 0

While the final momentum, after the astronaut throws the box, is:

p_f=mv+MV

where

m = 11 kg is the mass of the box

M = 79 kg is the mass of the astronaut

v = -0.45 m/s (to the left) is the velocity of the box (we take left as negative direction)

V is the final velocity of the astronaut

The total momentum is conserved, so

p_i = p_f\\0=mv+MV

And solving , we find V:

V=-\frac{mv}{M}=-\frac{(11)(-0.45)}{79}=0.06 m/s

And the positive sign indicates that the direction is to the right.

Learn more about momentum:

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6 0
3 years ago
Which one of the following lines best illustrates personification?
Mrrafil [7]
The answer would be A, because the wind cannot complain, therefore it has been given a human quality.
7 0
3 years ago
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A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of
grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

KE = \frac{1}{2} mv^2

Replacing with our values we have that

KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2

KE = 1.1935*10^{-13}J

Therefore the kinetic energy of the helium nucleus is 1.1935*10^{-13}J

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

KE = qV

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

V = \frac{KE}{q}

V = \frac{1.19*10^{-13}}{1.6*10^{-19}}

V = 743750eV

Therefore the kinetic energy in electron vols is 743750eV

PART C) Applying the same relationship but now using the Helium core load, we will have to

KE = QV

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

V = \frac{KE}{Q}

But we need to note that the charge is equal to the number of charge for the unit charge, then

Q = \text{No. Charge} \times \text{Unit Charge}

Q = (2)(1.6*10^{-19}C)

Q = 3.2*10^{-19}C

Now replacing we have that

V= \frac{1.19*10^{-13}}{32*10^{-19}}

V = 371875V

Therefore the voltage applied is  371875V

5 0
3 years ago
A small ferryboat is 3.90 m wide and 6.30 m long. when a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t
Leya [2.2K]
When the truck's weight is added to the boat, the boat sinks 5 cm deeper,
and displaces additional water whose weight is equal to the weight of the
truck.

The volume of the additional displaced water is

             (3.9 m) x (6.3 m) x (5.0 cm)

         =  (3.9 m) x (6.3 m) x (0.05 m)  =   1.2285 m³ .

The weight of that much water is the weight of the truck.

          Mass of 1 liter of water  =  1 kilogram

          1.2285 m³  =  1,228.5 liters  =  1,228.5 kg of water.

          Weight = (mass) x (gravity)

                      = (1,228.5 kg) x (9.8 m/s²)  =  12,039 Newtons.

                                                                  (about 2,708 pounds) 

6 0
3 years ago
A piston-cylinder arrangement contains air modeled as an ideal gas with constant specific heat ration of k =1.4. The air undergo
loris [4]

Answer:

Pish posh idk xD im busy but ill help in a second

Explanation:

4 0
2 years ago
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