All categories

3 years ago

Step 3 - Fillin the missing numbers: 21 -3 -10 24 b d

Physics

2 answers:

Studentka2010 [4]3 years ago

7 0

Answer:

b = 2

c = -1

d = -2

Complete the division. The remainder is 0.

The quotient is A. x2 - x - 12

Explanation:

on edg.... Good Luck!!!

mash [69]3 years ago

4 0

Answer:

b = 2

c = -1

d = -2

Explanation:

Got this question right on edge

You might be interested in

Examples of uniform velocity

slava [35]

Explanation:

When a truck travels in equal distances in equal intervals of time then we say that the body has got a uniform velocity. In the above example a truck is traveling at 5 miles in all the positions at A, B, and C and all in the intervals of 5 minutes each.

8 0

3 years ago

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

Inessa05 [86]

Explanation:

The given data is as follows.

Inner wheel Radius = 20 m,

Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

Angular velocity of automobile = w = $\frac{V}{R}$

where, V = linear velocity of automobile m/min,

R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

u = linear velocity of inner wheel

and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

w =

= $\frac{u}{(R - d)}$

u = $V \times \frac{(R - d)}{R}$

= $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus, u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

= $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

= $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =

= $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

= $\frac{V}{r} \times 0.1651$

5 0

3 years ago

Please help me I need this for a test

Marysya12 [62]

Answer:

Explanation:

Lol thats big tufffffffffvffffffftg

fuf

8 0

3 years ago

A man pushes on a wall what does Newton third law say must happen

Verdich [7]

If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.

8 0

3 years ago

Read 2 more answers

Roughly to what height would a 5 kg stone need to be raised in order to have the same stored energy as the energy stored in the

hoa [83]

8.16m is the required height, a 5kg stone need to be raised.

One sort of potential energy is gravitational potential energy, which is equal to the product of the object's mass (m), the gravitational acceleration (g), and the object's height (h) as measured in relation to the ground's surface (the body).

We obtain the formula by considering the work done in raising a mass m through a height h.

Work in elevating mass m through height h is equal to force times distance.

The force must be greater than the mass m's weight, hence F = mg.

Work done = mgh = gravitational potential energy

Energy = Mass of the object × gravitational acceleration × height.

Mass of the stone = 5kg

Equating ;

∴ 400 J = 5 kg × 9.8 m/s² × height

Height = 8.16 m

Therefore, 8.16m is the required height.

Learn more about energy here:

brainly.com/question/1242059

#SPJ1

8 0

2 years ago