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4 years ago

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A ramp is 2.8 m long and 1.2 m high. How much power is needed to push a box up the ramp in 4.6 s with a force of 96 N?

1 answer:

tino4ka555 [31]4 years ago

6 0

Power = work /time

and W = force * dis

W = 96 * 3.046 J = 292.44 J

P = 292.44 / 4.6 = 63.575 watt

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6.A Soldering Iron has a power of 25W. It uses the 230V mains supply. What is the current when it is switched on?

solmaris [256]

Power (P) in "watts" in a circuit is the product in multiplying the current (I) in "amperes" and voltage (V) in "volts".

P = I x V ; I = P / V

Substituting the given values,
I = (25 W) / (230 V) = 0.1087 amperes

Therefore, the current is approximately 0.1087 A.

4 0

4 years ago

The temperature of a 2.0kg block increases by 5C when 2000 J of thermal energy are added to the block. What is the specific heat

vitfil [10]

Explanation:

Q=mc(T2-T1)

or

q = mcΔT ,

where m is the mass of the sample,

c is the specific heat,

and ΔT is the temperature change.

Q=2.0 × 2000 × 5

Q=20000J⋅kg −1 ⋅K −1

I hoped I helped pls rate as brainliest

7 0

2 years ago

An experiment is performed aboard the International Space Station to verify that linear momentum is conserved during collisions

Anni [7]

Answer:

$v=15.9554\ m.s^{-1}$

Explanation:

Given:

mass of the honey drop 1, $m_1=35.5\times 10^{-3}\ kg$

velocity of the honey drop 1, $v_1=13.1\ m.s^{-1}$

mass of the honey drop 2, $m_1=52.3\times 10^{-3}\ kg$

velocity of the honey drop 2, $v_2=14.5\ m.s^{-1}$

mass of the honey drop 3, $m_1=75.7\times 10^{-3}\ kg$

velocity of the honey drop 3, $v_3=18.3\ m.s^{-1}$

<em>In ISS there is zero gravity an the collision is completely inelastic.</em>

<u>So, applying the law of conservation of momentum:</u>

$m_1.v_1+m_2.v_2+m_3.v_3=(m_1+m_2+m_3).v$

$35.5\times 10^{-3}\times 13.1+52.3\times 10^{-3}\times 14.5+75.7\times 10^{-3}\times 18.3=(35.5+52.3+75.7)\times 10^{-3}\times v$

$v=15.9554\ m.s^{-1}$

4 0

3 years ago

a 645 N person is wearing stiletto shoes. If she lifts her left leg, and rocks back onto the heel under her right leg, the press

fgiga [73]

Answer:

0.96cm

Explanation:

Given parameters:

Weight of person = 645N

Pressure of the heel = 9000000PA

Unknown

Diameter of the circular base = ?

Solution

Pressure is the force per unit area of a body.

Pressure = $\frac{Force}{Area}$

To solve this problem, find the unknown area of the circle;

Pressure x Area = Force

Area = $\frac{Force}{Pressure}$

Area = $\frac{645}{9000000}$ = 7.17 x 10⁻⁵m²

Note: 1Pa = 1Nm⁻²

Now;

Area of a circle = π r²

r = $\frac{d}{2}$

Area = π x $\frac{d^{2} }{4}$

d is the diameter

d² = 4 x Area x (1/π)

d² = 4 x 7.17 x 10⁻⁵ x (1/3.14) = 9.13 x 10⁻⁵

d = 0.0096m

To cm;

100cm = 1m

0.0096m; 0.0096 x 100= 0.96cm

7 0

3 years ago

A 77.0 kg rider sitting on a 7.3 kg bike is riding along at 9.3 m/s in the positive direction. The rider drags a foot on the gro

saw5 [17]

Answer with Explanation:

We are given that

Mass of rider=m=77 kg

Mass of bike =m'=7.3 kg

Initial velocity,u=9.3 m/s

Final velocity,v=6 m/s

A.Change in velocity=v-u=6-9.3=-3.3 m/s

Total mass,M=m+m'=77+7.3=84.3 kg

Change in momentum= $M(v-u)=84.3(-3.3)=-278.19 kgm/s$

B.Impulse=Ft=Change in momentum=-278.19kg m/s

C.Time,t=13.2 s

$v=u+at$

Using the formula

$6=9.3+13.2a$

$13.2a=6-9.3=-3.3$

$a=-\frac{3.3}{13.2}=-0.25 ms^{-2}$

F= $ma$

$F=84.3(-0.25)=-21.075 N$

D. $S=ut+\frac{1}{2}at^2$

$S=9.3(13.2)+\frac{1}{2}(-0.25)(13.2)^2$

$S=100.98 m$

5 0

4 years ago