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VARVARA [1.3K]
3 years ago
12

A custodian pushes a 119 kg box across a level floor. The box moves at a constant velocity. The custodian pushes horizontally on

the box with a force of 724 Newtons, what must be the friction force experienced by the box in Newtons?
Physics
1 answer:
bonufazy [111]3 years ago
8 0

The friction force experienced by the box in Newtons is 724N

The frictional force is the force that opposes the moving force acting on a body on a horizontal plane.

  • For a static body, the acceleration will be zero,

According to Newton's second law of motion:

  • ∑Fx = ma
  • +Fm - Ff = ma

Given the following paramters

  • Moving force Fm = 724N
  • acceleration a = 0m/s²
  • mass m = 119 kg

Substitute the given parameters into the formula to get the frictional force:

724 - Ff = 119(0)

724 - Ff = 0

Ff = 0 + 724

Ff = 724N

Hence the friction force experienced by the box in Newtons is 724N

Learn more here: brainly.com/question/8859573

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A stationary siren creates an 894 Hz
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Answer:

12.3 m/s

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

Given:

fs = 894 Hz

fr = 926 Hz

c = 343 m/s

vs = 0 m/s

Find: vr

926 = 894 (343 + vr) / (343 + 0)

vr = 12.3

The speed of the car is 12.3 m/s.

5 0
2 years ago
Rock composed of many thin layers. This image appears in an Earth science magazine with this caption: "A non-foliated rock found
blondinia [14]

Answer:

its B

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4 0
3 years ago
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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1
luda_lava [24]

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

4 0
2 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

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=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

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Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

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