1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
PSYCHO15rus [73]
3 years ago
6

And the density of water is 1000kgm3 plz urgent​

Physics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer: yes

Explanation:

You might be interested in
Starting from rest, a car takes 2.4s to travel 15m. Assuming a constant acceleration, how long will it take the car to travel th
Alborosie

Answer:

T = 2.4 + 2.4 = 4.8 [s]

Explanation:

In order to solve this problem, we must use the following kinematics equation and calculate the acceleration value.

x=x_{o} +v_{o}*t+(\frac{1}{2})*a*t^{2}

Vo = inital velocity = 0

x - xo = 15 [m]

t = time = 2.4 [s]

15 = 0.5*a*(2.4)^2

a = 5.208 [m/s^2]

We can use the same equation to find the time.

30 = 15 + 0.5*(5.208)*t^2

t = 2.4 [s]

T = 2.4 + 2.4 = 4.8 [s]

7 0
3 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

3 0
3 years ago
I need helpppppppppppppppppp!
allsm [11]

Answer:

i think its B

Explanation:

but check it before do it  

3 0
3 years ago
Read 2 more answers
When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?
Nezavi [6.7K]
We have F = kx or ma = kx where m and k are constants. Therefore, if x is halved, a must be halved too.
5 0
3 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
Other questions:
  • What is the difference in the charges on a ballon rubbed in your hair and a glass rod rubbed
    5·1 answer
  • During an experiment, a student moved a cell from pure water to salted water. What will most likely happen to the cell?
    10·1 answer
  • An atom has 25 protons 30 neutrons and 25 electrons what is the charge of the atom nucleus
    7·1 answer
  • A shear wave (S wave) is a type of seismic _________ that shakes the ground back and forth perpendicular to the direction the wa
    13·1 answer
  • The urethra extends from the:
    5·1 answer
  • Investigators are working on a case where they need to know whether a watch will stop when it is dropped. In order to have a ver
    14·2 answers
  • LOADS OF POINTS - PLEASE SHOW WORKING AND METHOD USED
    12·2 answers
  • 7) Se calcula dividiendo distancia entre tiempo
    12·1 answer
  • Which describes the innermost layer of Earth?​
    12·1 answer
  • How many atoms of carbon, C, are in 0.020 g of carbon?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!