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AleksAgata [21]

3 years ago

7

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

1 answer:

Mademuasel [1]3 years ago

4 0

Hope this answers your question!! Ask any help at anytime

You might be interested in

An object has more momentum if it has which of the following?

maks197457 [2]

Answer:

C.) A high velocity and Large mass.

Explanation:

Momentum of any object is defined by following formula

Here : m = mass of object

v = velocity of object

now we know that since momentum is product of mass and velocity

So in order to have more momentum we need the value of this product to be more. So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.

6 0

2 years ago

Narysuj wykres zależności szybkości od czasu i drogi od czasu jeśli ciało porusza się ruchem jednostajnym z szybkością 45 m/s.

murzikaleks [220]

Lett me come back imma translate this... and then ill come to help

7 0

3 years ago

Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What

Sauron [17]

Answer:

0.426 L

Explanation:

Boyles law is expressed as p1v1=p2v2 where

P1 is first pressure, v1 is first volume

P2 is second pressure, v2 is second volume.

Given information

P1=96 kPa, v1=0.45 l

P2=101.3 kpa

Unknown is v2

Making v2 the subject from Boyle's law

$v2=\frac {p1v1}{p2}$

Substituting the given values then

$v2=\frac {96*0.45}{101.3}=0.4264560710760l\approx 0.426 l$

Therefore, the volume is approximately 0.426 L

6 0

2 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

2 years ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

7 0

2 years ago