Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 m/s

Physics

1 answer:

strojnjashka [21]3 years ago

6 0

Answer:

The correct answer is t = 0.92s

Explanation:

Initial velocity v0 = 3.0 m/s

Displacement Δy = ?

Acceleration a = -9.8m/s2

Final velocity v = -6.0m/s

Time t=? Target unknown

We can use the kinematic formula missing Δy to solve for the target unknown t:

V=v0+at

We can rearrange the equation to solve to t:

V-v0=at

t= v-v0/a

Substituting the known value into the kinematic formula gives:

t= (-6.0m/s)-(3.0m/s)

————————————

-9.8m/s2

= -9m/s

—————-

-9.8m/s2

=0.92s

You might be interested in

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$