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MissTica
3 years ago
7

A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

Physics
1 answer:
Len [333]3 years ago
7 0

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object.  We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t  = 55 s

v2 = ?

We can use a kinematic equation to solve this problem.  We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

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Gametes are the cells used during sexual reproduction to produce a new individual organism.

Explanation:

The male gamete or sperm, is a smaller, mobile cell that meets up with the much larger and less mobile female gamete, egg or ova.

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Playing Vedio Games. :)
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The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object
Ann [662]

The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object  is 16.4 N.

<h3>What is frictional force?</h3>

Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.

The normal  force acting on the object of mass 4.2 Kg is N = mg

N = 4.2 Kg × 9.8 m/s² = 41.16 N

Frictional force = ц N

                         = 0.40 × 41.16 N

                         = 16.4 N.

Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N

To find more on frictional force, refer here:

brainly.com/question/1714663

#SPJ1

Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

3 0
1 year ago
Where should an object be placed in front of a concave mirror’s principal axis to form an image that is real, inverted, larger t
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3 years ago
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A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
3 years ago
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