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3 years ago

What do we mean when we describe forces as balanced?

1 answer:

3 0

Well, first, let's recognize that there is no such thing as a force that is
balanced or unbalanced.

When we say that a group of two or more forces is balanced, we mean
that when you add up all the magnitudes and directions of the forces, the
whole group adds up to zero, so they have the same net effect as if there
were no force at all.

You might be interested in

Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through

KATRIN_1 [288]

Answer:

Index of refraction

Explanation:

or refractive index

5 0

3 years ago

uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest

vfiekz [6]

Answer:

a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J

Explanation:

a) work is defined by

W = F . x

the bold letters indicate vectors, in this case the force is electric

F = q E

we substitute

F = q E x

the charge of the electron is

q = - e

F = - e E x

let's calculate

W = - 1.6 10⁻¹⁹ 365 3 10⁻²

W = - 1.752 10⁻¹⁸ J

b) the change in potential energy is

U = q ΔV

the potential difference is

ΔV = - E. Δs

we substitute

U = - q E Δs

the charge of the electron is

q = - e

U = e E Δs

we calculate

U = 1.6 10⁻¹⁹ 365 3 10⁻²

U = + 1.752 10⁻¹⁸ J

3 0

2 years ago

What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

4 0

3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

muminat

Answer:

7 meters, 2.8 meters

Explanation:

work done (nm) = force (n) * distance (m)

140= 20 * m

140/20 = m

m=7 meters

140= 50 * m

140/50 = m

m= 2.8 meters

4 0

2 years ago