All categories

3 years ago

angela uses a force of 25 newtons to lift her grocery bag while doing 50 joules of work. how far did she lift the grocery bags

Physics

1 answer:

Evgen [1.6K]3 years ago

5 0

We know that:
W=Fs
50J=25N*s
s=50J/25N
s=2m

You might be interested in

A centrifuge was used to separate into it’s components. What conclusions out the sample can be formed based on the technique use

Tatiana [17]

-- The sample was a fluid.

-- It was a mixture or a suspension ... NOT a solution.

4 0

3 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

3 years ago

Noㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

GREYUIT [131]

Answer:

?????...............

4 0

3 years ago

Read 2 more answers

Which of the following is a sign of a physical Change ?

guapka [62]

C.) i believe because the change in shape is changing the physical appearance

7 0

3 years ago

Read 2 more answers

How do you calculate the total resistance of a series circuit

CaHeK987 [17]

Add all the resistances across the circuit together the calculate the total resistance

5 0

3 years ago