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3 years ago

How does a higher heat capacity affect the amount of time it takes for an object’s temperature to change?

Physics

1 answer:

Alenkasestr [34]3 years ago

8 0

$higher~ specific ~heat ~capacity ~means ~ it ~has ~higher ~heat ~ energystorage$

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4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer

Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

3 0

3 years ago

What type of atom is K2CO3

My name is Ann [436]

<span>Potassium carbonate (K2C03) is white salt and is often </span>found damp. It is soluble in water which makes a strong concoction. Hope this helps.

4 0

2 years ago

Read 2 more answers

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

sattari [20]

I can't make sense of this question. Julie's throwing the ball, so it's leaving her rather than arriving at her ???

3 0

2 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

2 years ago

A kid applies a force of 70 N to a ball over 1.1 meters. How much work did he do on the ball?

LekaFEV [45]

Answer:

77J

Explanation:

Not really an explanation to this, I just had this lesson last year and remembered it.

Hope I helped! ☺

8 0

3 years ago