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3 years ago

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Kiley went 5.7 km/h north and went 5.8km/h west. From start to finish,she went 8.1 km/h northwest which best describes her numbe

rs?

1 answer:

ladessa [460]3 years ago

8 0

These values are all vectors since they all possess both magnitude and direction.

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What type of data can be used to make a model of food?

beks73 [17]

Answer:

Below:

Explanation:

Food balance sheet data are useful in monitoring trends in food consumption over time and in making rough comparisons between countries. Often, such data are the only data that can be readily obtained for the rapid evaluation of new problems.

Hope it helps...

It's Muska... :)

4 0

3 years ago

Read 2 more answers

How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules

dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0

3 years ago

What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m

earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒<em> v² = u² + 2as</em>, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

<em>The acceleration of the train must be - 0.133 m/s²</em>

3 0

3 years ago

In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

Ratling [72]

Answer:

Electric force, $F=-3.59\times 10^6\ N$

Explanation:

Given that,

Electric charge 1, $q_1=+40\ C$

Electric charge 2, $q_2=-40\ C$

Distance, $d=2\ km=2\times 10^3\ m$

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

$F=k\dfrac{q_1q_2}{d^2}$

k is the electrostatic constant

$F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}$

$F=-3.59\times 10^6\ N$

So, the electric force between these two sets of charges is $-3.59\times 10^6\ N$ .

5 0

4 years ago

You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w

solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

= 0.025×< -2, 0, -7 >

= < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

= 0.025×< 4, 4, -3 >

= < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact = < -0.05, 0, 0.175 > + < 0.1, 0.1, -0.075>

= < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity = v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

= < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0

3 years ago