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3 years ago

A weight lifter lifts a 1.0 x 102 kg mass 1.5 m in 2.0 s. The power is?

Physics

1 answer:

Serga [27]3 years ago

6 0

Answer:

$\boxed{p = {7.4 \times 10}^{2} W} \to \: option \: c.$

Explanation:

$his \: power \: is \to \\ p = \frac{mgh}{t} \\ p = \frac{1.0 { \times 10}^{2} \times 1.5 \times 9.8}{2.0} \\ p= 735 \\ \boxed{p = {7.4 \times 10}^{2} }$

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A car drives on a circular road of radius R. The distance driven by the car is given by = + [where a and b are constants, and t

pishuonlain [190]

Answer:

The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"

answers:

a. $18a m/s^{2}$

b. $a_{rad}=\frac{(27a +b)^{2}}{R}$

Explanation:

First let state the mathematical expression for the tangential acceleration and the radial acceleration.

a. tangential acceleration is express as

$a_{tan}=\frac{d|v|}{dt} \\$

since the distance is expressed as

$d=at^{3}+bt$

the derivative is the velocity, hence

$V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\$

hence when we take the drivative of the velocity we arrive at

$a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2$

b. the expression for the radial acceleration is expressed as

$a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}$

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3 years ago

What's going to happen in 3 weeks ?

Nana76 [90]

Answer:

god knows.

Explanation:

7 0

3 years ago

Read 2 more answers

You're driving in a car at 50 km/h and bump into a car ahead traveling at 48 km/h in the same direction. the speed of impact is

salantis [7]

To solve this problem, we must remember about the law of conservation of momentum. The initial momentum mist be equal to the final momentum, that is:

m1 v1 + m2 v2 = (m1 + m2) v’

where v’ is the speed of impact

Since we are not given the masses of each car m1 and m2, so let us assume that they are equal, such that:

m1 = m2 = m

Which makes the equation:

m v1 + m v2 = (2 m) v’

Cancelling m and substituting the v values:

50 + 48 = 2 v’

2 v’ = 98

v ‘ = 49 km/h

The speed of impact is 49 km/h.

6 0

3 years ago

Who was the first woman to compete and win a championship in the Olympics

natita [175]

Hélène de Pourtalès she was the first women

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3 years ago

A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,

zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

4 0

3 years ago