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Montano1993 [528]
3 years ago
9

A stone is dropped from a bridge abd it hits the water 2.2 seconds later how high is the bridge above the water

Physics
1 answer:
Bess [88]3 years ago
8 0

Answer:

h = 23.716 m

Explanation:

Given that,

The time taken by the stone to hit the water is, t = 2.2 s

Height of the bridge above the ground, h = ?

The distance that the body will fall through the time is given by the formula

                                S = 1/2 gt²  m

Where,        

                              g - acceleration due to gravity

Substituting the values in the above equation

                               S = 1/2 x 9.8 m/s² x (2.2 s)²

                                  = 23.716  m

Therefore, the height of the bridge from the surface of the water is h = 23.716  m

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Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,
yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0
3 years ago
The wavelength of 10^15-Hz light is:
Vanyuwa [196]

Answer:

D.300nm

Explanation:

Wavelength = Speed of light / Frequency of light.....

where the speed of light is...(3 × 10^8)

Wavelength = (3 × 10^8)/(1 × 10^15)

Wavelength = 3 × 10^-7

;Wavelength = 300 × 10^-9

Hence its....300 nm

5 0
2 years ago
Which two chemists organized elements based on properties such as how the elements react or whether they are solid or liquid
Lelechka [254]
Doberiner and lavoiser
5 0
2 years ago
Read 2 more answers
An isotope undergoes radioactive decay by emitting radiation that has a –1 charge. What other characteristic does the radiation
weeeeeb [17]

<u>Answer:</u> The radiation emitted will have negligible mass number.

<u>Explanation:</u>

Radioactive decay is defined as the process in which an unstable nuclei breaks down into stable nuclei via various methods.

An isotope undergoes a radioactive decay to attain stability.

There are three types of decay process, but the process in which the emitted radiation carries a charge of -1 is beta decay.

Beta decay is defined as the decay process in which a neutron gets converted to a proton and an electron.  In this decay process, beta particle is emitted. The emitted particle carries a charge of -1 units and has a mass of 0 units. The released beta particle is also known as electron.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

Hence, the radiation emitted will have negligible mass number.

3 0
3 years ago
Read 2 more answers
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
3 years ago
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