All categories

3 years ago

Cumulative Exam

Physics

2 answers:

Pani-rosa [81]3 years ago

8 0

Answer:

A. Plane mirrors produce only virtual images, and concave mirrors produce real and virtual images.

Explanation:

dalvyx [7]3 years ago

4 0

Answer:

Explanation:

Hi! It would be A. Plane mirrors produce only virtual images,and concave mirrors produce real and virtual images.

You might be interested in

Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a

dusya [7]

Answer:

$W=-21,870,000\ J$

Explanation:

Work and Kinetic Energy

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

$W=\Delta k=k_1-k_0$

Since

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

$\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}$

$W=0-21870000\ J$

$\boxed{W=-21,870,000\ J}$

3 0

3 years ago

A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will

gtnhenbr [62]

1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
$x(t)=v_0t$
$y(t)=h- \frac{1}{2}gt^2$
where the motion on the x-axis is a uniform motion with constant speed $v_0=18.1 m/s$ , while the motion on the y-axis is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ , and initial height $h=1.85 m$ .

First of all, we can find the time t at which the ball reaches the ground by requiring $y(t)=0$ :
$0=h- \frac{1}{2}gt^2$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s$

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
$x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m$

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
$v_x=18.1 m/s$
On the y-axis, the velocity is given by
$v_y(t)=gt$
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
$v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s$
And the speed of the ball is the magnitude of the resultant of the two components:
$v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s$

8 0

3 years ago

Read 2 more answers

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

mass of the block, = M
coefficient of static friction in section 1, = $\mu_s_1$
angle of inclination of the plane, = θ

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How

kirill115 [55]

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

$E_{pot}-E_{friction}=E_{kin}$

where:

$E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}$

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

$(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]$

4 0

3 years ago

Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30ms

nexus9112 [7]

S= ?
U= 0 $ms^{-1}$
V= 30 $ms^{-1}$
A= 9.8 $ms^{-2}$
T=

$V^2 = U^2 + 2AS$
$900 = 19.6S$
$\frac{900}{19.6} = S$
S = 45.92 $m$

6 0

4 years ago