A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?

A 1.00 kg block of ice, at -25.0°C, is warmed by 35 kJ of energy. What is the final temperature of the ice?

ahrayia [7]

Answer:

-8.4°C

Explanation:

From the principle of heat capacity.

The heat sustain by an object is given as;

H = m× c× (T2-T1)

Where H is heat transferred

m is mass of substance

T2-T1 is the temperature change from starting to final temperature T2.

c- is the specific heat capacity of ice .

Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.

Hence ; 35= 1× c× ( T2-(-25))

35= c× ( T2+25)

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}

T2= 16.60-25= -8.40°C

C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}

6 0

2 years ago

You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of

solong [7]

Answer:

The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

V₂₁ = V₂ -V₁

∴ V₂ = V₂₁ + V₁

On substitution

V₂ = 3 + 2

= 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0

3 years ago

An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o

borishaifa [10]

Hi there!

Recall the equation for centripetal force:
$F_c = \frac{mv^2}{r}$

We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
$r * F_c = mv^2$

Divide both sides by Fc:
$\boxed{ r= \frac{mv^2}{F_c}}$

7 0

2 years ago

A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the sepa

AnnZ [28]

Answer:

Increases

Explanation:

The expression for the capacitance is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.

The expression for the energy stored in the parallel plate capacitor is as follows;

$E=\frac{Q^{2}}{2C}$

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.

Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.

Therefore, the option (c) is correct.

8 0

3 years ago

Andy has two strong magnetic marbles of equal size. What will happen when he puts the two marbles about 2 centimeters apart?

VikaD [51]

I believe it would be C

5 0

3 years ago

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