Ask question

Ask question

All categories

2 years ago

7

The acceleration due to gravity for any object, including 1 washer on the string, is always assumed to be m/s2. The mass of 3 wa

shers, when converted to kg, is kg. The applied force of 3 washers will increase the applied force on the car to N.

2 answers:

Andrews [41]2 years ago

6 0

Answer: 10, 0.0147, 0.147

Explanation: Correct on edge2020

Pachacha [2.7K]2 years ago

5 0

Answer:

The force will increase in proportion to the mass of the objects

Explanation:

The acceleration due to gravity is always the same. It is expressed in meters per second squared or m/s². The figure of 9.81 m/s² is an average value that was taken after calculating the acceleration under different surfaces. In fact, the acceleration differs depending on the shape of the part of the earth in relation to the earth's magnetic field and force.

Thus, if one washer was 20 kg, the acceleration being 9.81 m/s² the weight will be:

F = ma

= $20 * 9.81\\= 196.2 N$

If there are there washers, the weight will be:

F = 3 * 20 * 9.81

= 588.6 N

You might be interested in

In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being

german

Answer: Got It!

<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.

6 0

2 years ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

#For the proton:

$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

7 0

2 years ago

Two resistors have resistances R1= ( 24+- 0.5)ohm and R2 = (8 +-0.3)ohm . Calculate the absolute error and the percentage relati

zysi [14]

Answer:

Absolute error=0.006

Percentage Relative error=0.6

Explanation:

The resistors have resistance of 24 ohm and 8 ohm.

The change in resistance is 0.5 and 0.3 ohm respectively.

Relative error for parallel combination of resistors is

= dR/R²

= dR1/R1² + dR2/R2²

= 0.5/(24)² + 0.3/(8)²

= 0.5/ 24*24 + 0.3/ 64

= 0.5/576+0.3/ 64

= 32 + 172.8/ 36,864

=204.8/ 36,864=0.0055

=0.006

Percentage error =Relative error *100

= 0.006* 100 = 0.6

8 0

2 years ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

2 years ago

Which is a correct representation of .000025 in scientific notation?

lidiya [134]

0.000025 → 2.5 × 10⁻⁵ → 2.5E-5

<h3>Further explanation</h3>

Scientific notation represents the precise way scientists handle exceptionally abundant digits or extremely inadequate numbers in the product of a decimal form of number and powers of ten. Put differently, such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent or power. It is a system for expressing extremely broad or exceedingly narrow digits compactly.

Scientific notation should be in the form of

$\boxed{ \ a \times 10^n \ }$

where

$\boxed{ \ 1 \leq a \ < 10 \ }$

The number 'a' is called 'mantissa' and 'n' the order of magnitude.

From the key question that is being asked, we face the standard form of 0.000025.

$\boxed{ \ 0.000025 = \frac{25}{1,000,000} \ }$

The coefficient (or mantissa), i.e. 25, is still outside of 1 ≤ a < 10. Both the numerator and denominator are divided by 10.

$\boxed{ \ 0.000025 = \frac{2.5}{100,000} \ }$

The denominator consists precisely of five zero digits.

Hence, 0,000025 is written in scientific notation as $\boxed{\boxed{ \ 2.5 \times 10^{-5} \ or \ 2.5E - 5 \ }}$

The inverse of scientific notation is the standard form. To promptly change scientific notation into standard form, we reverse the process, move the decimal point to the right or left. This expanded form is called the standard form.

<u>A notable example:</u>

$\boxed{ \ 3.0 \times 10^{8} \ Hz \ \rightarrow 300,000,000 \ Hz \ or \ 300 \ MHz}$

<h3>Learn more</h3>

0.00069 written in scientific notation brainly.com/question/7263463
Express the pill’s mass in 0.0005 grams using scientific notation or in milligrams brainly.com/question/493592
What 3 digits are in the units period of 4,083,817 brainly.com/question/558692

Keywords: which is, a correct representation, 0.000025, in scientific notation, expanded form, exponent, base, standard form, mantissa, the order of magnitude, power, decimal, very large, small, figures, abundant digits, inadequate

6 0

2 years ago

Read 2 more answers