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zhenek [66]
2 years ago
5

If a car is accelerating downhill under a net force 3674 N , what additional force would cause the car to have a constant veloci

ty
Physics
1 answer:
Semenov [28]2 years ago
6 0

Answer:

For the car to move with constant velocity the additional force required is  F__{dg} }=  -3674 \  N

Explanation:

From the question we are told that

  The net force of the car is  F_{net} = 3674 \  N

Generally the total  force acting on the  car is the net force plus the force due to gravity acting in direction of the car (Let denote it as F__{dg})

So the total force acting on the car is mathematically represented as

        F_{net} + F__{dg}} = F

Here this F representing the total force can be mathematically represented as

      F =  ma

Now for constant velocity to be attained, the acceleration of the car will be zero  

So  at constant velocity

       F =  m * 0

=>    F =   0 \  N

So  

     F_{net} + F__{dg}} = 0

=>   F__{dg} }=  -F_{net}

=>   F__{dg} }=  -3674 \  N

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Answer:

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and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

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2 years ago
A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

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How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

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H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

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10 months ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

Initial temperature, Ti = 205°

Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

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8 0
2 years ago
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