2 years ago

you run 2.5m/s horizontally from the edge of a cliff and 3 sec . later hit the ground, how high is the cliff?

Physics

1 answer:

bonufazy [111]2 years ago

3 0

The cliff would be 7.5m high

You might be interested in

Garret's swimming coach is giving him important and helpful tips to improve his swimming skills. Which gesture suggests that Gar

bogdanovich [222]

Answer: "important and helpful" Show garret is paying close attention to her coach. Hope this help!

7 0

3 years ago

Read 2 more answers

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

My name is Ann [436]

Answer:

915m

Hope this helps.

5 0

3 years ago

Suppose an endothermic reaction has a positive change in entropy greater than the heat absorbed. What can be said about the spon

pentagon [3]

<span>If the entropy is greater than the enthalpy, it will have more spontinaity</span>

8 0

3 years ago

Read 2 more answers

A Honda Civic Hybrid weighs about 3200 lb .

Usimov [2.4K]

3200÷0.22= 145.4545...N
(it is an infinite decimal)

3 0

3 years ago

Read 2 more answers

A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

$B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

$B=6.978*10^{-5}T$

Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

$a=\frac{qvB}{m}$

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

8 0

2 years ago