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Setler [38]
3 years ago
7

Answer and mark you as brainliest it's that easy

Physics
2 answers:
Tanzania [10]3 years ago
7 0
The answer is a because i said so
Oduvanchick [21]3 years ago
3 0

Answer:

a)

t=\frac{v-u}{a}

b) 35m/s

c) 0

d) -35m/s

e) Velocity is decreasing.

f) 70 seconds

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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
A block of mass 500g is pulled from rest on ahorizontal frictionless bench by a steady force F and travels 8m in 2s find the acc
Korvikt [17]

<u>We are Given:</u>

Mass of the block (m) = 500 grams or 0.5 Kg

Initial velocity of the block (u) = 0 m/s

Distance travelled by the block (s) = 8 m

Time taken to cover 8 m (t)= 2 seconds

Acceleration of the block (a) = a m/s²

<u>Solving for the acceleration:</u>

From the seconds equation of motion:

s = ut + 1/2* (at²)

<em>replacing the variables</em>

8 = (0)(2) + 1/2(a)(2)²

8 = 2a

a = 4 m/s²

Therefore, the acceleration of the block is 4 m/s²

3 0
3 years ago
If the mass of an object i if the mass of an object is 44 kg in its velocity is 10 m/s East how much kinetic energy
vampirchik [111]
K=0.5 mu×u
K=2200J no matter the direction
8 0
4 years ago
Why do individuals have different traits
Gwar [14]
Individuals have different traits because they are not twins so they should not  be alike also  because people have different genes witch would give them different traits. For example not every individual has blonde hair and blue eyes because they did not inherit those traits.
3 0
3 years ago
A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol
olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa,  т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P )  = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅  ,   where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅  of  CD on the shaft BC

Torque = 660 * 0.3 cos∅

             = 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0
3 years ago
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