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svp [43]
3 years ago
14

Lisa is conducting an investigation to determine the coefficient of friction between two surfaces. She uses a 25 kg block. What

is this object's weight, in newtons?
Physics
2 answers:
artcher [175]3 years ago
7 0

Answer:

Weight, W = 245 N

Explanation:

It is given that,

Mass of the block, m = 25 kg

We need to find its weight. From Newton's second law of motion,

F=m\times a

a is the acceleration of the object

On the surface of earth, a=9.8\ m/s^2

F=25\ kg\times 9.8\ m/s^2                    

F = 245 N

So, the weight of the object is 245 N. Hence, this is the required solution.

Paladinen [302]3 years ago
4 0

Answer:

245.25

Explanation:

1 kg is 9.81 newtons

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What term was used to describe the final digit
xxMikexx [17]

Answer:

Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement. A number with more significant digits is more precise. For example, 8.00 cm is more precise than 8.0 cm.

7 0
3 years ago
IF YOU GOOD AT SCIENCE PLEASE ANSWER THIS ASAP
xz_007 [3.2K]

Answer:

I would believe that it would be the last option

Explanation:

Physical science is a type of science that mainly focuses on natural objects that are not alive, such as minerals and rocks.

4 0
2 years ago
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The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of
jasenka [17]

1) Frequency: 3.29\cdot 10^{15}Hz

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

E=2.18 aJ=2.18\cdot 10^{-18} J

The energy of a photon is given by

E=hf

where h=6.63\cdot 10^{-34}Js is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz


2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

\lambda=\frac{c}{f}

where c=3\cdot 10^8 m/s is the speed of light and f is the frequency. Substituting the frequency, we find

\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm

5 0
3 years ago
Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free
sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

           ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

           ΔE = W

internal energy can be related to the change in temperature

            ΔE = 3/2 n K ΔT

we substitute

           3/2 n (T₂-T₁) = W

as the work is on the gas it is negative

            W = 3/2 n (T₁- T₂)

3 0
3 years ago
A large 10.kg medicine ball is caught by a 70.kg student on the track team. If the ball was moving at 4.0 m/s, how fast will the
exis [7]

v2 = ?

m1 = 10kg

m2 = 70kg

v1 = 4m/s

E1 = E2

E1 = 1/2 * m1 * v1^2 = 1/2 * 10kg * 4m/s^2 = 80J

E2 = 1/2 * m2 * v2^2 = 80 J

v2 = √(E2/(2 * m2)) = √(80J/(2 * 70kg)) = about 0.76m/s

7 0
3 years ago
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