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masya89 [10]
2 years ago
15

What horizontally-applied force will accelerate a crate of mass 400 kg at 1 meter per second per second across a factory floor a

gainst a friction force of half its weight?
Physics
1 answer:
son4ous [18]2 years ago
6 0

Answer:

The horizontally applied force = 2360 N

Explanation:

<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)

Fh = Fr + ma......... Equation 1

Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.

<em>Given: m = 400 kg, a = 1 m/s²</em>

Fr = 1/2 W, W = mg  ⇒W = 400×9.8 = 3920 N

∴Fr = 1/2(3920), Fr = 1960 N

Substituting these values into equation 1

Fh = 1960 + 400×1

Fh = 1960 + 400

Fh = 2360 N

Therefore the horizontally applied force = 2360 N

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A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
IrinaK [193]

Answer : The specific heat capacity of the alloy 1.422J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of alloy = ?

C_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of alloy = 21.6 g

m_2 = mass of water = 50.0 g

T_f = final temperature of system = 31.10^oC

T_1 = initial temperature of alloy = 93.00^oC

T_2 = initial temperature of water = 22.00^oC

Now put all the given values in the above formula, we get

21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC

c_1=1.422J/g^oC

Therefore, the specific heat capacity of the alloy 1.422J/g^oC

6 0
2 years ago
*NEED ANSWER STAT*
tresset_1 [31]
Becomes a +1 ion for this
6 0
3 years ago
Read 2 more answers
A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?
Andreas93 [3]

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation \frac{distance}{time}. After substitution, this gives the fraction \frac{320 000}{14 400}, which reduces to 22 \frac{2}{9} m/s, or about 22.2 m/s.

4 0
3 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

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Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
you give a shopping cart a job down the aisle the cart is full of groceries and has a mass of 18 kilograms the cart accelerates
mezya [45]

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As long as you continue pushing the cart with 54 newtons of force,
it will accelerate at 3 m/s². 

At the instant you release it, or keep your hands on it but stop pushing,
it will stop accelerating.  It'll continue forward at the speed it had when
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2 years ago
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