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schepotkina [342]
2 years ago
14

Please help with this

Physics
1 answer:
sertanlavr [38]2 years ago
6 0

Answer:I have to say 56

Explanation: because it is going up by 8

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PLS I NEED HELP ASAAP​
Lerok [7]

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

x=t-t^2

so, x_0  means when time is zero so we replace t with zero in the equation,

x_0=(0)-(0)^2\\x_0=0

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

v=1-2t\\

now for  v_0  we insert t = 0 and get

v_0=1-2(0)\\v_0=1

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

a=-2

so Option A is your answer.

Remember derivative of a constant is always zero because a constant value has no rate of change has its a constant hence the derivative is 0

5 0
2 years ago
Answer the question below based on the periodic table entry for bromine.
Sphinxa [80]
The answer is 45 i hope i could help
8 0
3 years ago
Read 2 more answers
Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou
belka [17]
The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year. 
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather
4 0
3 years ago
Point P and point charge Q are separated by a distance R. The electric field at point P has magnitude E. How could the magnitude
Angelina_Jolie [31]
<span>We know , E = kQ/r^2 where q = charge and r is separation between point and point charge. Now, At P, E= kQ/r^2 Since, Q can't be changed, we can do that by varying r 2E = 2kq/r^2 2E = kq/ (r/ sqrt2)^2 Hence, if we bring Q closer such that distance between P and Q becomes r/ sqrt 2, E will get doubled.</span>
5 0
3 years ago
A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision
Nimfa-mama [501]

Answer:

Δ KE =  249158.6 kJ  

Explanation:

given data

Truck mass  M =  1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum  that is express as

Mu = (M+m) V     .......................1

put here  value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy  will be here

Δ KE =   \frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2         .................2

put here value and we get

 Δ KE = \frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2  

solve it we get

Δ KE =  249158.6 kJ  

6 0
3 years ago
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