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3 years ago

PLEASE ANSWER QUICK!! A student pushes a wagon full of bricks with a constant force across the ground. Which of

Physics

1 answer:

givi [52]3 years ago

5 0

Answer:

Explanation:

If you do B, the wagon slows down due to friction

If you do C, it would slow down the wagon by half

If you do D, it would slow it down, due to there being more weight but same force.

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A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

OverLord2011 [107]

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

$\theta_i = 34 degree$

so we know that

$\theta_i = \theta_r$

so here we can say that reflection angle of light will be same as angle of incidence

$\theta_r = 34 degree$

8 0

3 years ago

If a planet has a radius 20% greater than that of the Earth but has the same mass as the Earth, what is the acceleration due to

Vaselesa [24]

Answer:

$g_2=6.8125 m^2/sec$

Explanation:

We know that

Acceleration due to gravity g is given by the formula

$g= \frac{GM}{r^2}$

G= gravitational constant

M= mass of the Earth

r= radius of the earth

$g_1 = GM/r_1^2$

Let acc. due to gravity after radius is 20% greater be g_2

then

$g_2=GM/r_2^2$

=> g1/g2 = (r_2/r_1)^2 => g2 = 9.81/1.2^2 = 6.8125

5 0

2 years ago

Read 2 more answers

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

7 0

3 years ago

1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package

Sidana [21]

The velocity of the package after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

and t is time

To calculate the velocity of the package after it has fallen for 3.0 s

That means, we will determine the value of v, at time t = 3.0 s

The parameters are

u = 0 m/s

g = 9.8 m/s²

t = 3.0 s

Putting these values into the equation

v = u + gt

We get

v = 0 + (9.8×3.0)

v = 0 + 29.4

v = 29.4 m/s

Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s

Learn more here: brainly.com/question/13327816

6 0

2 years ago

Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector

garri49 [273]

Answer:

$5.2\ \text{m/s}$

$70^{\circ}$ south of east

Explanation:

$v_a$ = 3 m/s

$\theta_a$ = $20^{\circ}$ north of east

$v_b$ = 6 m/s

$\theta_b$ = $40^{\circ}$ south of east = $360-40=320^{\circ}$ north of east

x and y component of $v_a$

$v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}$

$v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}$

x and y component of $v_b$

$v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}$

$v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}$

$\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}$

Magnitude

$|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}$

Direction

$\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}$

The magnitude of the change in velocity vector is $5.2\ \text{m/s}$ and the direction is $70^{\circ}$ south of east.

6 0

3 years ago