1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Cerrena [4.2K]
3 years ago
5

A student drops a rock from a bridge to the water 12m below. How many seconds does it take the rock to hit the water?

Physics
1 answer:
ANEK [815]3 years ago
7 0
R = \frac{g t^{2} }{2}
r - displacement or height
t - time taken
g = 10 m/s-2
t = square root from 2r/g = \sqrt{ \frac{2 x 12}{10} } = 1.5 seconds
You might be interested in
In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
2 years ago
Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water
artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0
3 years ago
Force F → = (−8.0 N)iˆ + (6.0 N)jˆ acts on a particle with position vector r → = (3.0 m)iˆ + (4.0 m)jˆ. What are (a) the torque
disa [49]

Answer with Explanation:

We are given that

F=-8\hat{i}+6\hat{j}

r=3\hat{i}+4\hat{j}

a.We have to find the torque on the particle about the origin.

We know that

Torque=\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}

By using the formula

\tau=50\hat{k}

b.\mid \tau\mid =\mid F\mid \mid r\mid sin\theta

\mid F\mid=\sqrt{(-8)^2+(6)^2}=10

\mid r\mid=\sqrt{3^2+4^2}=5

\mid \tau\mid=\sqrt{(-50)^2}=50

Substitute the values then we get

50=10\times 5 sin\theta

sin\theta=\frac{50}{50}=1

sin\theta=sin90^{\circ}

Because sin90^{\circ}=1

\theta=90^{\circ}

3 0
3 years ago
How much power does it take to do 500 J of work in 10 seconds?
Dmitry_Shevchenko [17]
Power = work/time
  
          = 500/10
 
          = 50J/s or 50 watt 


7 0
3 years ago
What activity is mainly a strength training excerise?
Sever21 [200]

Biceps curls & pushups , benchpress

6 0
3 years ago
Other questions:
  • Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a
    8·1 answer
  • The weight of the block in the drawing is 97.0 N. The coefficient of static friction between the block and the vertical wall is
    15·1 answer
  • Please help me with my physics I don’t understand! Please provide work.
    8·1 answer
  • What mass of water at 23.0°C must be allowed to come to thermal equilibrium with a 1.75-kg cube of aluminum initially at 150°C t
    12·2 answers
  • Stars located near _____ appears to move in circles when viewed from earth
    11·1 answer
  • Our brain works to create:
    8·2 answers
  • suppose a block of lead of mass 0.4kg at temperature of 95°C is dropped in a 2 kg of water originally at 20°C in a container. wh
    6·1 answer
  • When should scientific claims be questioned?
    6·1 answer
  • Which vector is the sum of the vectors shown below ?
    5·1 answer
  • A thin rod of length d on a frictionless surface is pivoted about one end
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!