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yuradex [85]
2 years ago
4

hree boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in

the center. A force of 50 N pushes on the 5.0-kg box, which pushes against the other two boxes. What magnitude force does the 5.0-kg box exert on the 3.0-kg box?
Physics
1 answer:
Maslowich2 years ago
6 0

Answer:

25 N

Explanation:

We are given that three boxes rest side-by-side on a smooth, horizontal floor.

m_1=5 kg

m_2= 3 kg

m_3=2 kg

Force applied on 5 kg box=50 N

We have to find the magnitude force exert on the 3 kg box by 5 kg box.

According to given question

50=(5+3+2)a

a=\frac{50}{10}=5 m/s^2

When 5 kg box exert force on 3 kg box then equal and opposite force exert on 5 kg box due to 3 kg box and 2 kg box.

Let F be the force exert by 5 kg box on 3 kg box and

50-F=5\cdot 5

F=50-25=25 N

Hence, the magnitude force exert on 3 kg box by 5 kg box=25 N

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Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s
exis [7]

Answer

given,

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

  Δ P = m Δ v                  

  \Delta P = m (v_f-v_i)

  \Delta P = 1300 (0 - 11)

  Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

    I = - 1.43 x 10⁴ kg.m/s

c) magnitude of force experienced by Kara

  I = F x t

 I is impulse acting on the car

 t is time

  - 1.43 x 10⁴= F x 0.14

    F = -1.021 x 10⁵ N

negative sign represents the direction of force

8 0
2 years ago
Read 2 more answers
A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus
dexar [7]

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

6 0
2 years ago
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
2 years ago
Read 2 more answers
Calculate the pressure, in atmospheres, exerted by each of the following:
gregori [183]

Answer:

a) 14.2 atm

b) 4.46 atm

c) 1.06 atm

Explanation:

For an ideal gas,

PV = nRT

P = pressure of the gas

V = volume occupied by the gas

n = number of moles of the gas

R = molar gas constant = 0.08206 L.atm/mol.K

T = temperature of the gas in Kelvin

a) For HF,

P =?, V = 2.5L, n = 1.35 moles, T = 320K

P = 1.35 × 0.08206 × 320/2.5

P = 14.2 atm

b) For NO₂

P =?, V = 4.75L, n = 0.86 moles, T = 300K

P = 0.86 × 0.08206 × 300/4.75

P = 4.46 atm

c) For CO₂

P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K

P = 2.15 × 0.08206 × 330/55

P = 1.06 atm

4 0
2 years ago
An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .
miss Akunina [59]

Answer:

The airplane should release the parcel 6.7*10^2 m before reaching the island

Explanation:

The height of the plane is y_0=100m, and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

y=y_0 - \frac{gt^2}{2}    [1]

And the distance X is

x = V.t     [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

X = V . \sqrt{\frac{2y_0}{g}}

Using the given values:

x = 150 m/s  \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}

x = 6.7*10^2 m

4 0
2 years ago
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