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ExtremeBDS [4]
3 years ago
8

Lolidk what this is​

Physics
2 answers:
alex41 [277]3 years ago
3 0

Answer:

I know

Explanation:

Physics= hard

natita [175]3 years ago
3 0
Lol‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️‍♀️
You might be interested in
A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0
3 years ago
B. Jerome plays middle linebacker for South's varsity football team. In a game against
weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0
3 years ago
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
2 years ago
When using a spring scale, the measurements you obtain will be in ____.
SpyIntel [72]
The answer is letter c

6 0
3 years ago
Read 2 more answers
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
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