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user100 [1]
3 years ago
7

34.1 grams / 1.1 mL =

Physics
1 answer:
nataly862011 [7]3 years ago
6 0

Answer:

31,000kg/m^3

Explanation:

Hopefully this helps

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this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85
mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}

(b)

\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}

(c)

\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}

3 0
1 year ago
True or false? An object at rest has an instantaneous acceleration of zero.
Tpy6a [65]
True. it’s at rest which means it’s not moving so there’s no acceleration
8 0
3 years ago
Select True or False for the following statements about conductors in electrostatic equilibrium. All points of a conductor are a
erik [133]

Answers :

1. All points of a conductor are at the same potential. - True

2. Charges prefer to be uniformly distributed throughout the volume of a conductor. - False

3 The electric field inside the conducting material is always zero. -True

4.Just outside the surface of a conductor, the electric field is always zero. - False

4 0
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
0.0884 moles of a diatomic gas
Sloan [31]

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

\Delta\ U = nC_{v}\Delta\ T

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J

Now, we can apply First Law of Thermodynamics as follows:

\Delta\ Q = \Delta\ U + W

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\

<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>

7 0
3 years ago
Read 2 more answers
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