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2 years ago

2. How can we make the percent error value as minimal as possible?

Physics

2 answers:

andriy [413]2 years ago

8 0

Answer:

Percent error can be reduced by improving both your accuracy and precision.

Molodets [167]2 years ago

3 0

Answer:

Percent error can be reduced by improving both your accuracy and precision.

Explanation:

You might be interested in

A 210 kg meteoroid is heading toward the earth accelerates by 2.4 × 105 m/s2.

lidiya [134]

According to Newton's second law of motion, Force is the product of mass and acceleration of the object.
So, F = m * a

Here, m = 210 Kg
a = 2.4 * 10⁵ m/s²

Substitute their values,
F = 210 * 2.4 * 10⁵ N
F = 504 * 10⁵ N
F = 5.04 * 10⁷ N

In short, Your Answer would be Option B

Hope this helps!

6 0

2 years ago

A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?

Varvara68 [4.7K]

Answer:

Explanation:

A force of $55\text{ }N$ is acting on a wagon along the road. The wagon weights $7.5\text{ }kg$ . Acceleration of the wagon is given as $5.3\text{ }\frac{m}{s^{2}}$ .

Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

Gravitational Force and Normal Reaction cancel out each other.

Net External Force = Mass of system/wagon $\times$ Acceleration of wagon

$F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N$

$F_{friction}$ has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0

2 years ago

A sailboat weighing 980 lb with its occupants is running downwind at 8 mi/h when its spinnaker is raised to increase its speed.

Effectus [21]

Answer:

78.498N

Explanation:

The Net force provided by the spinnaker can be obtained from Newton's second law of motion as follows;

$F=\frac{m(v-u)}{t}................(1)$

where m is the mass, v is the final velocity, u is the initial velocity and t is the time interval for which the force acted.

Given;

m =980lb

v = 12mi/h

u =8mi/hr

t = 10s.

It is important to convert all quantities to their SI units where necessary, so we do that as follows;

1lb = 0.45kg,

hence 980lb = 980 x 0.45kg = 441kg.

1mile = 1609.34m

1hour = 3600s,

therefore;

$8mi/h=\frac{8*1609.34m}{3600s}=3.58m/s$

$12mi/h=\frac{12*1609.34m}{3600s}=5.36m/s$

Substituting all values into equation (1), we obtain the following;

$F=\frac{441(5.36-3.58)}{10}\\F=\frac{441*1.78}{10}\\F=\frac{784.98}{10}\\F=78.498N$

4 0

2 years ago

Ray Of Light [21]

Acceleration = Force \ mass

0,375N/0,60kg=0.6ms-2

4 0

1 year ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

2 years ago