All categories

3 years ago

Rosa uses the formula (vi cos)t to do a calculation. Which value is Rosa most likely trying to find?

Physics

2 answers:

Llana [10]3 years ago

8 0

Answer;

C - the horizontal displacement of a projectile launched at an angle

Explanation;

-The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally (vix) and the amount of time (t) that it has been moving horizontally.

-The range (or horizontal displacement) will increase as the angle is increased from 0 degrees to 45 degrees. The maximum range occurs at 45 degrees. As the angle is further increased to values greater than 45 degrees, the horizontal displacement decreases.

kaheart [24]3 years ago

6 0

The best and most correct answer among the choices provided by the question is <span>the horizontal component acceleration of a projectile launched at an angle </span>
I hope my answer has come to your help. God bless and have a nice day ahead!

You might be interested in

A geothermal plant has been built in a location. Which of these is the most likely benefit of the geothermal plant in this locat

kramer

Answer:

Explanation:

A and B are not true and D is a disadvantage

4 0

2 years ago

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k

Brrunno [24]

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$

Explanation:

The expression for the gravitational force is

$F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}$

Differentiate the above equation

$\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}$

The velocity is the distance in at time so

$V=\frac{dr}{dt}=0.4 \frac{km}{s}$

$\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}$

$\frac{dF}{dt}=-8\frac{N}{s}$

8 0

3 years ago

Nucleus A decays into the stable nucleus B with a half-life of 22.07 s. At t=0 s there are 1,293 A nuclei and no B nuclei. At wh

Alexeev081 [22]

Answer:

29.38 seconds

Explanation:

Half life, T = 22.07 s

No = 1293

Let N be the number of atoms left after time t

N = 1293 - 779 = 514

By the use of law of radioactivity

$N=N_{0}e^{-\lambda t}$

Where, λ is the decay constant

λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second

so,

$514=1293e^{-0.0314t}$

$2.5155=e^{0.0314t}$

take natural log on both the sides

0.9225 = 0.0314 t

t = 29.38 seconds

5 0

2 years ago

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If th

Korvikt [17]

Answer:

A(3.56m)

Explanation:

We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.

We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.

Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2

Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).

Ef= -mgh+3/4mv^2

Since Ef=Ei=0

Mgh=3/4mv^2

gh=3/4v^2

h=0.75v^2/g

plug in givens to get h= 3.57m

5 0

3 years ago

An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti

shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

$I=F\Delta t$

where:

I is the impulse

F is the force exerted on the object

$\Delta t$ is the time during which the force is applied

For the object in this problem, we have

$F=106 N$ (force applied)

$\Delta t= 15.6 s$ (time interval)

Therefore, the impulse experienced by the object is:

$I=(106)(15.6)=1654 kg m/s$

3 0

2 years ago