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4 years ago

A scientist would use a balance to measure

Physics

2 answers:

GrogVix [38]4 years ago

7 0

Answer is B

Hope it Helps!

Art [367]4 years ago

7 0

A Balance is used to measure mass

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A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

nignag [31]

Answer:

Explanation:

Given

mass of box $m=80 kg$

coefficient of kinetic friction $\mu _k=0.15$

coefficient of Static friction $\mu _s=0.30$

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

$f_r=ma$

where $f_r=frictional\ Force$

$a=acceleration$

$a=0$

$f_r=0$

4 0

3 years ago

How much potential energy does a 40-N medicine ball gain when it is lifted 5 m?

777dan777 [17]

We know, Potential Energy = Force * Height
Here, F = 40 N
h = 5 m

Substitute their values,
U = 40 * 5
U = 200 J

In short, Your Answer would be Option A

Hope this helps!

6 0

3 years ago

If anyone anwers this question, i will give you 100 points and brainliest!

Tju [1.3M]

Answer:

33. at the top?

34. to add more "energy" to the water and keep it moving

35. the dam because the energy is is building up

Explanation:

i tink that's correct sorry if not

have a good rest of ur day

3 0

3 years ago

Read 2 more answers

A cat is moving at 18 m/s when it accelerates for 2 seconds. What is its acceleration?

vova2212 [387]

Answer: acceleration: $4 m/s^{2}$

velocity: $26 m/s$

Explanation:

The complete question is written bellow:

A cat is moving at 18 m/s when it accelerates at $4 m/s^{2}$ for 2 seconds. What is his new velocity?

In this situation the following equation will be useful:

$V=V_{o}+at$

Where:

$V$ is the cat’s final velocity (new velocity)

$V_{o}=18 m/s$ is the cat’s initial velocity

$a=4 m/s^{2}$ is the cat's acceleration

$t=2 s$ is the time

Solving the equation:

$V=18 m/s+(4 m/s^{2})(2 s)$

$V=26 m/s$ This is the cat's new velocity

5 0

3 years ago

Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce

ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

$F=k\frac{q_{1}q_{2} }{r^{2} }$

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

$F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}$

4 0

4 years ago