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Crank
3 years ago
5

Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?

Physics
1 answer:
ozzi3 years ago
6 0

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

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A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. the dynamic coefficient of friction between the
Talja [164]
The first thing you should know is that the work is defined as:
 W = F * d
 Where
 F = force
 d = displacement
 We have then
 (a) the block
 F = (0.2) * (100) = 20
 d = 100
 W = (20) * (100) = 2000 ft.lbf
 (b) the man as the system.
 F = (0.2) * (100 + 180) = 56
 d = 100
 W = (56) * (100) = 5600 ft.lbf
 answer:
 (a) 2000 ft.lbf
 (b) 5600 ft.lbf
3 0
3 years ago
In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0
3 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

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       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

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

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

4 0
3 years ago
Which statement is correct? When a positively charged atom looses an electron to a positively charged atom, two neutral atoms ar
vazorg [7]
I believe the answer is "When a neutral atom looses an electron to another neutral atom, two charged atoms are created."
3 0
3 years ago
Read 2 more answers
A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t
Semmy [17]

Answer:

T = 40.501\,^{\textdegree}C

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

-Q_{out,Cu} = Q_{in,H_{2}O}

After a quick substitution, the expanded expression is:

-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)

-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot  (T-39^{\textdegree}C)

43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T

The final temperature of the system is:

T = 40.501\,^{\textdegree}C

8 0
3 years ago
Read 2 more answers
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