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lyudmila [28]
3 years ago
8

What is a controlled experiment? none of the variables are changed all variables are kept constant two variables are changed at

a time one variable is changed at a time
Physics
1 answer:
Ostrovityanka [42]3 years ago
7 0

One variable is changed at a time

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A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from
Alla [95]

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

6 0
3 years ago
an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque
Misha Larkins [42]

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0
3 years ago
How many infrared photons of frequency 2.57 x 1013 Hz would need to be absorbed simultaneously by a tightly bound molecule to br
victus00 [196]

Answer:

94

Explanation:

f = 2.57 x 10^13 Hz

E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J

Energy of each photon = h f

Where, h is Plank's constant

Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J

Number of photons = Total energy / energy of one photon

N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94

6 0
3 years ago
A pool and stops at the
Bas_tet [7]

In Linear motion the swimmer swims

8 0
2 years ago
During a snowball fight two balls with masses of 0.4 kg and 0.6 kg, respectively, are thrown in such a manner that they meet hea
valkas [14]
Considering conservation of momentum;
m1v1 + m2v2 = m3v3

In which,
m1 = mass of snowball 1 = 0.4 kg
v1 = velocity of snowball 1 = 15 m/s
m2 = mass of snowball 2 = 0.6 kg
v2 = velocity of snow ball 2 = 15 m/s
m3 = combined mass = 1 kg
v3 = velocity after comination
Therefore;
0.4*15 + 0.6*15 = 1*v3
v3 = 6+9 = 15 m/s
KE = 1/2mv^2

Then,
KE1 = 1/2*0.4*15^2 = 45 J
KE2 = 1/2*0.6*15^2 = 67.5 J
KE3 = 1/2*1*15^2 = 112.5 J

Therefore, KE3 (kinetic energy after collision) = K1+K2 {kinetic energy before collision). And thus it is 100%.
3 0
3 years ago
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