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zhuklara [117]
2 years ago
7

What is the net force for this? (please don't forget it's direction and magnitude)​

Physics
1 answer:
avanturin [10]2 years ago
3 0

Answer:

Explanation:

As the sum of the two right directed forces match exactly the left directed force, the only unbalanced force, and thus the net force, is the upward 25 N force.

You might be interested in
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H
Reika [66]

Answer:

The answer to your question is:

a)  t1 = 2.99 s ≈ 3 s

b)  vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ?   time to reach the ground

vf = ?   final speed

a)    h = vot + (1/2)gt²

     74 = 10t + (1/2)9.81t²

     4.9t² + 10t -74 = 0                  solve by using quadratic formula

   

   t = (-b ± √ (b² -4ac) / 2a

   t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

   t = (-10 ± √ 1550.4 ) / 9.81

  t1 = (-10 + √ 1550.4 ) / 9.81               t2 = (-10 - √ 1550.4 ) / 9.81

  t1 = (-10 ± 39.38 ) / 9.81                    t2 = (-10 - 39.38) / 9.81

   t1 = 2.99 s ≈ 3 s                               t2 = is negative then is wrong there are

                                                                   no negative times.

b) Formula vf = vo + gt

                  vf = 10 + (9.81)(3)

                  vf = 10 + 29.43

                  vf = 39.43 m/s

4 0
2 years ago
Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D i
kkurt [141]

The change in potential along a path from C to D due to a small charged sphere is 900 V.

Given:

Charge, Q = 3 nC = 3 × 10⁻⁹ C

Distance between the sphere and point C, r₁ = 0.02 m

Distance between the sphere and point D, r₂ = 0.06 m

Calculation:

We know that the electric potential is given as:

V = k Q/r   - (1)

where, V is the electric potential

            k is Coulomb's force constant

            Qis the charge on the  sphere

            r is the  separation distance

The electric potential at point C due to charged sphere can be given as:

V₁ = k Q/r₁

   = (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]

   = 1350 V

The electric potential at point D due to charged sphere can be given as:

V₂ = k Q/r₂

   = (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]

   = 450 V

Now, the change in potential along the path from C to D can be calculated as:

ΔV = V₂ - V₁

     = 450 V - 1350 V

     = -900 V

The negative sign indicates that the work is done against the electric field in moving the charge from C to D.

Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.

Learn more about the electric potential here:

<u>brainly.com/question/12645463</u>

#SPJ4

8 0
1 year ago
Read 2 more answers
When reading a buret, where is the initial and final volumes taken from? The top (where the zero is) or the bottom?. If the bure
liubo4ka [24]
<span>When reading a buret, the initial reading should be taken from the top of the glassware and the final volume should still taken at the top. If the buret is completely, the initial volume for most buret would be zero. though, there are some where their initial starts at 50 decreasing to zero.</span>
5 0
3 years ago
Pls help on this one?
sattari [20]
The answer is point C
4 0
3 years ago
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