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3 years ago

NEED HELP PLEASE !!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

Anna [14]3 years ago

3 0

Answer:

(1) Exercise should be painful in order for it to be beneficial. FALSE

(2) Proper progression involves a gradual increase in the level of exercise. TRUE

(3) Exercising on a regular, consistent basis provides you with the best results. TRUE

Explanation:

(1) It is a false statement because whenever we do exercise, it should not be painful, it should be comfortable to some extent for a person, if it will be painful then it means one cant get the true and genuine benefit from the exercise.

(2) It is very much true because one must keep on increasing the level of his/her exercise gradually with the passage of time, it will make his/her body synchronized with the exercise properly.

(3) It is also a true statement because, one must use variety of exercises in his/her workout session targeting different muscles and different areas and parts of the body to make exercise really beneficial.

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A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =

stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

$d_{13} =\sqrt{24.68}$ * 10⁻²m = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

F₂₃x = F₂₃ = +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

$F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2} }$

$F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2} }$ *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0

3 years ago

For a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass 1000 kg. the truck then

babunello [35]

ANSWER

Both trucks will move together with speed v = 6.67 m/s

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

EXPLANATION

As we know that there is no external force on the system of two trucks

So here momentum of the two trucks before collision and after collision will remain same

So here we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

so here we will have

$v_{1i} = 10 m/s$

$v_{2i} = 0$

$m_1 = 2000 kg$

$m_2 = 1000 kg$

now we will have

$2000\times 10 + 1000\times 0 = (2000 + 1000) v$

$v = \frac{20000}{3000} = 6.67 m/s$

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

3 0

2 years ago

Read 2 more answers

What is center of gravity and how does it affect rockets?

Bezzdna [24]

Answer:

Rocket Center of Gravity. As a rocket flies through the air, it both translates and rotates. The rotation occurs about a point called the center of gravity. The center of gravity is the average location of the weight of the rocket.

4 0

3 years ago

Read 2 more answers

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

3 years ago

A nuclear power plant operates at 40.0% efficiency with a continuous production of 1192 MW of usable power in 1.00 year and cons

Maurinko [17]

Answer:

the answer is in the picture below......

4 0

3 years ago