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3 years ago

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Physics

1 answer:

hammer [34]3 years ago

4 0

Answer:

(a) J = 10560 kg-m/s (b) 251.42 N

Explanation:

Given that,

Mass, m = 4400 kg

Initial speed, u = 5.2 m/s

Final speed, v = 7.6 m/s

Time, t = 42 s

(a) Let J be the impulse exerted on the vehicle. Impuse is equal to the change in momentum such as :

J = m(v-u)

J = 4400 (7.6-5.2)

J = 10560 kg-m/s

(b) Impulse = Force × t

$F=\dfrac{J}{t}\\\\F=\dfrac{10560}{42}\\F=251.42\ N$

Hence, this is the required solution.

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When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st

defon

Answer:

Current in the hair dryer will be equal to 15 A

Explanation:

We have given that household is operated at 110 volt

So potential difference V =110 volt

Power drawn by hairdryer is P = 1650 watt

We have to find the current in the hair dryer

We know that power is given as P = VI, here V is potential difference and I is current

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4 years ago

How many electrons must be added to an object to give it an overall net charge of -3.0 x 10^-7 C? The charge of an electron is -

katovenus [111]

The number of electrons is $1.9\cdot 10^{12}$

Explanation:

The total net charge on a negatively-charged object is given by

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where:

N is the number of excess electrons

$e$ is the charge of a single electron

In this problem, we have

$Q=-3.0\cdot 10^{-7}C$ is the overall charge of the object

$e=-1.602\cdot 10^{-19}C$ is the charge of one electron

Solving the equation for N, we find:

$N=\frac{Q}{e}=\frac{-3.0\cdot 10^{-7}}{-1.602\cdot 10^{-19}}=1.9\cdot 10^{12}$

Learn more about electrons and other particles:

brainly.com/question/2757829

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3 years ago

A clay pot at room temperature is placed in a kiln, and the pot\'s temperature doubles. how much more heat per second is the pot

xeze [42]

Answer:

Explanation:

If we treat the pot as a black body, then:

q = σ T⁴ A,

where q is the heat per second radiated,

σ is the Stefan-Boltzmann Constant,

T is the absolute temperature,

and A is the surface area.

If the absolute temperature doubles, then q increases by a factor of 2⁴ = 16.

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3 years ago

A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio

pickupchik [31]

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

$m=\dfrac{W}{g}$

$m=\dfrac{110\ N}{9.8\ m/s^2}$

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

$M=d\times V$

d is the density of water, $d=1000\ kg/m^3$

V is the volume of rock, $V=0.00337\ m^3$

$M=1000\ kg/m^3\times 0.00337\ m^3$

M = 3.37 kg

The apparent weight in water, W = m - M

$W=7.85\ kg\times 9.8\ m/s^2$

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3 years ago

Calculate the current if a charge of 5.60C passes through a point in a conductor in 15.4s

Ulleksa [173]

Electric current is defined as charge per time in units of Coulombs/second
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7 0

3 years ago