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2 years ago

22. A ball is thrown horizontally from the roof of a building 12 m tall with a speed of 3.1 m/s.

Physics

1 answer:

zysi [14]2 years ago

5 0

Answer:

a) t = 1.6 s

b) d = 4.9 m

c) v = 16 m/s

d) θ = 79°

Explanation:

time of fall

t = √(2h/g) = √(2(12)/9.8) = 1.5649... s

d = vt = 3.1(1.56) = 4.8512...

vertical velocity vy = at = 9.8(1.56) = 15.336... m/s

v = √(15.336² + 3.1²) = 15.6464... m/s

θ = arctan(15.336/3.1) = 78.5724...°

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How much force is needed to accelerate a 68 kilogram skier at a rate of 1.2m/sec

xxMikexx [17]

Let's use Newton's 2nd law of motion:

Force = (mass) x (acceleration)

Force = (68 kg) x (1.2 m/s²) = 81.6 newtons .

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2 years ago

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A ferry coming into port is sailing at 12 m/s. It takes 2.5km to come to rest in the port. Calculate the deceleration of the fer

pogonyaev

Answer: -0.0288 m/s^2

Explanation:

Let's suppose that the ferry decelerates at a constant rate A (deceleration is an acceleration in the opposite direction to the original motion of an object)

Then the acceleration equation of the ferry will be:

a(t) = -A

(the negative sign is because this acceleration is in the opposite direction with respect to the movement of the ferry)

To get the velocity equation of the ferry, we need to integrate with respect to the time, t, we will get:

v(t) = -A*t + v0

where v0 is the initial velocity of the ferry, v0 = 12m/s.

v(t) = -A*t + 12m/s

For the position equation of the ferry we need to integrate again over time:

p(t) = (-A/2)*t^2 + (12m/s)*t + p0

Where p0 is the initial position of the ferry, in this case, it can be zero, because it will depend on where we put the origin on our coordinate axis.

then p0 = 0m

P(t) = (-A/2)*t^2 + (12m/s)*t

The ferry will come to rest at the moment when it's velocity is equal to zero, this will happen when:

v(t) = 0m/s = -A*t + 12m/s

We need to find the value of t.

A*t = 12m/s

t = (12m/s)/A

Now we can replace this in the position equation because we know that the ferry needs 2.5 km or 2500 meters to come to rest.

p( (12m/s)/A) = 2500m = (-A/2)*( (12m/s)/A)^2 + (12m/s)*((12m/s)/A)

2500m = (-72 m^2/s^2)/A + (144m^2/s^2)/A

2500m = (72 m^2/s^2)/A

2500m*A = (72 m^2/s^2)

A = (72 m^2/s^2)/2500m = 0.0288 m/s^2

and the acceleration of the ferry was -A, then the acceleration of the ferry is:

-0.0288 m/s^2

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2 years ago

An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an

liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = $\frac{I^2}{R}$ ........1

put here value we get

Power consumed = $\frac{16.7^2}{7.2}$

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

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3 years ago

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A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

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3 years ago

The magnetic field of a long, straight, and closely-wound solenoid, inside the solenoid at a point near the center, is 0.645 T.

Savatey [412]

Answer:

B'=1.935 T

Explanation:

Given that

magnetic field ,B= 0.645 T

We know that magnetic filed in the solenoid is given as

$B=\mu _0 n\ I$

I=Current

n=Number of turn per unit length

μ0 =magnetic permeability

Now when the current increased by 3 factors

I'=3 I

Then the magnetic filed

$B'=\mu _0 n\ I'$

$B'=\mu _0 n\ (3I)$

B'=3 B

That is why

B' = 3 x 0.645 T

B'=1.935 T

Therefore the new magnetic filed will be 1.935 T.

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3 years ago