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2 years ago

Convert 350 calories to KJ

Physics

1 answer:

Lesechka [4]2 years ago

5 0

If you convert 350 calories to kilojoules you will get 1.4644

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Which of the following is most like an indicator of a chemical change

qaws [65]

Answer:

methyl orange, methyl red,phenoptalin, merhy red

Explanation:

all this following are indicators use to check the end point of a reaction

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3 years ago

A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained

mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

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2 years ago

How do I calculate velocity

Jobisdone [24]

Answer: See photo

Explanation: There are a couple of ways to use velocity in an equation in the photo.

5 0

2 years ago

1. A large turbine has an initial angular momentum of 6700 kgm^2/s. A storm is rolling in and the wind picks up. 8 seconds later

LekaFEV [45]

Answer:

262.5 Nm

Explanation:

Torque is the rate of change of angular momentum.

Hence, we have

$\tau = \dfrac{\Delta L}{t}$

Δ<em>L</em> is the change in angular momentum.

Using values in the question,

$\tau = \dfrac{8800-6700 \text{ kg m}^2\text{/s}}{8\text{ s}} = 262.5 \text{ Nm}$

4 0

3 years ago

A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi

Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2$

$u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}$

$u=-2.496\ m/s$

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

$v = u+gt$

Put the value in the equation

$-2.496=0-9.8\times t$

$t =\dfrac{2.496}{9.8}$

$t=0.254\ sec$

The total time is

$t'=t+1.95$

$t'=0.254+1.95$

$t'=2.204\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times9.8\times(2.204)^2$

$s=23.80\ m$

Hence, (a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

4 0

2 years ago