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Vikentia [17]
3 years ago
8

Can someone please help me with these physics problems? I just don’t even know where to start.

Physics
1 answer:
KIM [24]3 years ago
3 0

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

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A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially a
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Answer:

M1 V1 = M1 V2 + M2 V3    conservation of momentum

V2 = (M1 V1 - M2 V3) / M1      where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9  =   (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

7 0
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What does it mean to say something is proportional?
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4 0
2 years ago
Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird
serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr  and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

  1. db = 2dr
  2. L + (L - x) = 2x
  3. 2L - x = 2x
  4. 2L = 3x
  5. x = \frac{2}{3}L

Insert it in x = \frac{2}{3}L

\frac{2}{3}(2.4km) = 1.6km

Now we use formula db = 2dr

  1. db = 2L - x
  2. db = 2(2.4km) - 1.6km
  3. <u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

  1. Vr = L/ Δt ⇒ Δt = \frac{L}{Vr}
  2. \frac{2.4km}{6.8km/hr}
  3. \frac{6}{17}hr

(Km cancel each other)

  1. Vb = db/Δt ⇒ db = VbΔt
  2. 13.6km/hr(\frac{6}{17}hr )
  3. <u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

6 0
3 years ago
4.
Akimi4 [234]

Answer:

c

Explanation:

plz make me brainliest i have answered

5 0
3 years ago
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