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2 years ago

Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity

Physics

1 answer:

Crazy boy [7]2 years ago

4 0

Answer:

The distance is

Explanation:

Apply the equation of motion

(

)

The initial velocity is

−

The acceleration is

−

Therefore, when

, we get

(

)

⋅

and when

(

)

⋅

Therefore,

The distance travelled in the fourth second is

(

)

−

(

)

−

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Question 3 of 10

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Answer:

D. Newton's second law

Explanation:

Newton's second law of motion states that force of an object is a product of its mass and its acceleration.

Mathematically, F= ma where m is mass and a is acceleration

So from the statement above : The acceleration of an object is proportional to the force applied to it and inversely proportional to its mass , it can be seen from the formula variation as;

F= ma -----making a the subject of the formula

a= F/ m

a= 1/m * F --------- a is inversely related to m as you can see from 1/m but directly related to F hence;

Increase in mass with the same force applied causes the body to accelerate slower where as when force increases, the body accelerates faster.

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2 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

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Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

In this case we have an electron (-e) and a proton (e), so:

$F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N$

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

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3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

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3 years ago

How much energy will a stock tank heater rated at 1790 Watts use in a 24 hour period? 1. 1790 × 24 × 3600 Joules 2. 1790 × 24 ×

Rashid [163]

<h2>Option 1 is the correct answer.</h2>

Explanation:

Power of heater, P = 1790 W

Time used, t = 24 hours = 24 x 60 x 60 = 24 x 3600 s

We have the equation

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}$

We need to find energy,

Substituting

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}\\\\1790=\frac{\texttt{Energy}}{24\times 3600}$

Energy = 1790 x 24 x 3600 J

Option 1 is the correct answer.

3 0

3 years ago

Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity​

Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity