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EastWind [94]
3 years ago
14

With an average acceleration of −1.3 m/s2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.0 m/s

to a complete stop?
Answer in units of s.
Physics
1 answer:
nevsk [136]3 years ago
3 0
13.0÷1.69
= seconds for the cyclist to completely stop

first solve -1.3 to the second power or -1.3 squared by multiplying -1.3 by itself
((-1.3)×(-1.3 )=1.69)
then divide 13 by 1.69
we are dividing because we are solving a deceleration equation and by multiplying 2 negatives the sum will always equal a positive and if we multiplied 13 by 1.69 we would turn this into an accelleration equation.

so divide 13 by 1.69
(13÷1.69=7.6923076923)
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In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other e
tester [92]

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle  and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

4 0
4 years ago
How are atomic emission spectra like fingerprints for the elements
storchak [24]

Atomic emission spectra are like fingerprints for the elements, because it can show the number of orbits in that elements as well as the energy levels of that element. As each emission of atomic spectra is unique, it is the fingerprint of element.

<u>Explanation: </u>

Each element has unique arrangement of electrons in different energy levels or orbits. So depending upon the difference in energy of the orbital, the emission spectra will be varying for each element. As the binding energy and excitation energy is not common for any two elements, so the spectra obtained when those excited electrons will release energy to ground state will also be unique.

As in atomic emission spectra, the incident light will be absorbed by the electrons of those elements making the electron to excite, then the excited electron will return to ground state on emission of radiation of energy. Thus, this energy of emission is equal to the difference between the energy of initial and final orbital. So the spectra will act like fingerprints for elements.

8 0
3 years ago
A horizontal spring required a force of 1.0 N to compress it 0.1 m. How much work is required to stretch the spring 0.4 m?
Crank

Answer:

<h2>0.5J</h2>

Explanation:

given data

Force applied F= 1N

extension e= 0.1m

let us find the spring constant first

applying

F=ke

k=F/e

k=1/0.1

k=10N/m

Step two:

Required is the work done

we know that the expression/formula for the work done by a spring is given as

Wd=1/2kx^2

x=0.4m

substitute

Wd= 1/2*10*0.4^2

Wd=0.5*10*0.16

Wd=0.5J

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