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julia-pushkina [17]
3 years ago
13

What effect does increasing mass have on the amount of friction generated

Physics
1 answer:
LekaFEV [45]3 years ago
4 0
The greater the mass, the greater the fiction generated
You might be interested in
Please help if you can
denis23 [38]

Answer:

The answer is B

Explanation:

Because when the both sides aren't balanced one side has to cause motion. (fall down)

6 0
2 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

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

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
A wave moving along a rope has a
Gennadij [26K]

Answer:

V = wavelength * frequency = 1.5 * 5.5 = 8.25 m/s

7 0
2 years ago
Calculate the net force on particle q1.
antoniya [11.8K]

By applying Coulomb's law between the charges, the net force on the charged particle q₁ due to particle q₂ and q₃ is -9.86 N.

<h3>Distance between q₂ and q₃</h3>

The distance between the second charge and the third charge is given as;

r = 0.3 m

<h3>Force on q₂ due to q₃</h3>

F_{2} = \frac{kq_2q_3}{r^2} \\\\F_{2}= \frac{(8.99\times10^9) (7.7 \times 10^{-6})(5.9\times 10^{-6}) }{0.3^2} \\\\F_2 = 4.54\ N

<h3>Net force on particle q₁</h3>

The net force on particle q₁ is determined by summing the individual forces together;

F(net) = F₁ + F₂

F(net) = -14.4 + 4.54

F(net) = -9.86 N

Thus, by applying Coulomb's law between the charges, the net force on the charged particle q₁ due to particle q₂ and q₃ is -9.86 N.

Learn more about Coulomb's law here: brainly.com/question/24743340

7 0
2 years ago
Meghan explains to Zach the difference between commensalism, mutualism and parasitism. Which sentence(s) did she include in her
vlabodo [156]
C is the correct answer I’m sure
8 0
3 years ago
Read 2 more answers
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