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jekas [21]
3 years ago
11

You apply a 56.3N force to a 4.2Kg cart towards the right. The cart accelerates towards the right at a rate of 0.8m/s2. What is

the magnitude, in Newtons, of the net frictional force (the sum of the only other forces, besides the 56.3N one you apply)? Please enter only a numerical answer, no units necessary.
Physics
1 answer:
vovangra [49]3 years ago
6 0

Answer:

52,94N

Explanation:

Frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. The main purpose of a frictional force is to create resistance to the motion of one surface over the other surface.

Calculation:

Le the value of frictional force be Fk, therefore acceleration a = (F- Fk)/m

where :

F is the applied force

m the mass of the cart

a = (F- Fk)/m = 0.8

F - Fk = 0.8m

Fk = F - 0.8m

Fk = 56.3 - (0.8×4.2)

Fk = 52.94 N

The magnitude, in Newtons, of the net frictional force is therefore 52.94N

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Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at
miv72 [106K]

Answer:

The angle at which the boat must head is - 22.47^{\circ}

Solution:

As per the solution:

Distance between the parallel banks, d = 40 m

The maximum speed of water, v' = 3 m/s

constant speed, u' = 5 m/s

Also,

The speed of water of the river at a distance of 'x' units from the  west bank is given as a sine function:

f(x) = 3sin(\frac{\pi x}{40})          (2)

Now, to determine the angel at which the boat must head:

The velocity of the engine of the boat:

v = u'cos\theta\hat{i} + u'sin\theta\hat{j}

v = 5cos\theta \hat{i} + 5sin\theta\hat{j}

The abscissa of the boat at time t:

v = 5cos\theta t\hat{i}

Now, from above and eqn(1) , we can write:

f(5cos\theta t) = 3sin(\frac{\pi \times 5cos\theta t}{40})

Now, boat's velocity at time t:

v = 5cos\theta \hat{i} + (5sin\theta + 3sin(\frac{\pi \times 5cos\theta t}{40})\hat{j}

In order to obtain the position of the boat, we integrate both the sides, we get:

r = 5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j} + C             (3)

Now, at r = 0:

0 = 5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j} + C

C = \frac{24}{\pi cos\theta}\hat{j}

Now, from eqn (3)

r = 5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40} + \frac{24}{\pi cos\theta})\hat{j}                      (4)

the baot will reach the point at y = 0 and x = 40

Now,

40 = 5cos\theta t

t = \frac{8}{cos\theta}

Substituting the above value of 't' in eqn (4):

r = 5sin\theta \frac{8}{cos\theta}\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta \frac{8}{cos\theta}}{40} + \frac{24}{\pi cos\theta})\hat{j}

We get:

48 + 40\pi sin\theta = 0

\theta = sin^{- 1}(\frac{- 48}{40\pi}) = - 22.47^{\circ}

8 0
2 years ago
Why can’t contour lines cross?
JulsSmile [24]

ANSWER B: A location can only have one elevation.

5 0
3 years ago
Caleb is investigating the effect of friction on the motion of an object. He uses the following supplies for the investigation:
Annette [7]
Part a
Place the ramps on the floor and time each ramp for time it takes for wooden block too reach thee floor. The times will be different and thus you can conclude that friction will increase the time
Part b
Caleb uses the same block as this is his control variable that he wants to keep same so that the results are correct and accurate
Part c
The independent variable is the thing you change before you test so it will be the surface on which the wooden block is sliding
The dependent variable is the variable you measure so that will be the time it takes for the block to slide to the bottom.
7 0
3 years ago
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 8.00 ✕ 106 m/s pe
KonstantinChe [14]

Answer:

4347.8 m/s  

Explanation:

It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

\frac{m_ev_e^2}{r}=qv_eB\\\frac{m_pv_p^2}{r}=qv_pB\\

Take the ratio:

m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_e}{m_p}v_e\\\Rightarrow v_p=\frac{1}{1840}\times 8.0\times 10^6 m/s\\\Rightarrow v_p=4347.8m/s

3 0
3 years ago
WILL GIVE BRAINLIEST!
yaroslaw [1]

stable equilibrium, if displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement.

unstable equilibrium, if displaced it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly.

neutral equilibrium, is when an equilibrium is independent of displacements from its original position.

Have a good day, hope this helps

8 0
3 years ago
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