3 years ago

Which function represents g(x), a reflection of f(x) = 6(one-third) Superscript x across the y-axis?

Physics

1 answer:

Alisiya [41]3 years ago

6 0

Answer:

g(x)=6(3)°x

$g(x) = 6(3){x}$

You might be interested in

Power is expressed in what.?

TiliK225 [7]

Yes, that's correct.

5 0

3 years ago

Convert 27,549 into scientific notation

dezoksy [38]

2.7549 x 10^4 is the answer I hope this helped u

7 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

5 0

3 years ago

Answer the following questions

zaharov [31]

Answer:

9 - 10N to the left

10 - There is no change on the object

Explanation:

Can I have brainliest answer pls?

7 0

3 years ago

Read 2 more answers

The free body diagram four objects are shown Which object has a net force of -5 N down?

max2010maxim [7]

Answer:

F-5-89

Explanation:

7 0

3 years ago