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Mkey [24]
3 years ago
13

I need help please...

Physics
2 answers:
Natalka [10]3 years ago
6 0
No cluuuueee :/ sowwwwyyy but good luck
fomenos3 years ago
4 0
No wayyyyyyyyyyyyyyyyyy idk
You might be interested in
Which of the following was an effect of the French Revolution?
lozanna [386]

Answer: C. the elimination of special privileges for members of the first and second estates

Explanation:

The French Revolution was a period of political change in France. The French Revolution led to the proclamation of the first French Republic, radical political and social change, creation of constitutional monarchy, formation of the French consulate etc.

From the options given, the effect of the French Revolution was the the elimination of special privileges for members of the first and second estates. Therefore, the correct option is C.

5 0
3 years ago
Will mark brainliest. TRUE OR FALSE. the acceleration due to gravity is caused by a field force
choli [55]

Answer:

Yes, It is true.

The acceleration due to gravity is caused by field force.

5 0
2 years ago
A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?
murzikaleks [220]
The answer is:  " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V  ;  That is,  "mass divided by volume" ;
 
Density is expressed as:
__________________________________________    
                   "mass per unit volume";  in which the "mass" is expressed in units of "g" ("grams") ;  and the "unit volume" is expressed in units of:
    "cm³ " or "mL"; 
_____________________________________________
           {Note the exact equivalent:  1 cm³ = 1 mL }.
____________________________________________
         →  The formula is:  " D = m / V "  ; 
___________________________________________
   in which:

     "D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given); 

     "m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
                 
     "V" refers to the "volume", in units of "cm³ " ;
               which is:  "23.4 cm³ " (given);
_________________________________________________
                 We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________ 
              D  =  m / V ; 
_________________________________________________________
             And we rearrange; to isolate "m" (mass) on ONE side of the    equation; and then we plug in our known/given values;
 to solve for "m" (mass);  in units of "g" (grams) ;
___________________________________________________
    Multiply each side of the equation by "V" ; 
____________________________________________________
             V * { D  =  m / V } ;  to get:
____________________________________________________
      V * D = m ;   ↔   m = V * D ;
___________________________________________________
           Now, we plug in the given values for "V" (volume) and "D" (density) ;     to solve for the mass, "m" ;
______________________________________________________
           m  =  V * D ;
 
           m  =  (23.4 cm³) * (8.9 g / 1 cm³)  = (23.4 * 8.9) g = 208.26 g ;
  
 →  Round to "208 g" (3 significant figures);  
____________________________________
The answer is:  " 208 g " .
_____________________________________________________
7 0
3 years ago
May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
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