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2 years ago

13

A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work i

s done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules

1 answer:

podryga [215]2 years ago

6 0

The answer is 3) 480 joules

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A charge of 1. 5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How

ELEN [110]

It is a type capacitor is in which two metal plates arranged in such a way so that they are connected in parallel . The potential energy is stored in the capacitor will be 2.7×10⁻⁵.

<h3>What is parallel plate capacitor ?</h3>

It is an type capacitor is an in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.

A dielectric medium is a must in between these plates. helps to stop the flow of electric current through it due to its non-conductive nature .

The given data in the problem is;

Q is the charge= 1.5 µC

V is the change in voltage across the plates is = 36 V.

U is the potential energy=?

The formula for the potential energy is given by;

$\rm U= \frac{1}{2} \times Q \times V \\\\ \rm U= \frac{1}{2} \times 1.5\times 10^{-6} \times 36 \\\\ \rm U=2.7\times10^{-5}$

Hence the potential energy is stored in the capacitor will be 2.7×10⁻⁵.

To learn more about the parallel plate capacitor refer to the link;

brainly.com/question/12883102

3 0

1 year ago

Experiment: Group 1 drinks 500 mL of coffee a day.

slega [8]

Answer:

How does the drink content affect an individual's blood pressure?

Explanation:

In every experiment using the scientific method, an observation lays the foundation of that experiment. A problem must be observed, which then leads to asking a SCIENTIFIC QUESTION in order to investigate. A scientific question must include the variable being changed called INDEPENDENT VARIABLE and the variable being measured called DEPENDENT VARIABLE.

In this experimental procedure or set up,

- Group 1 drinks 500 mL of coffee a day.

- Group 2 drink 500 mL of tea a day,

- Group 3 is a control group i.e no drink

At the end of 60 days all participants

blood pressure is tested.

This set up indicates that the variable being changed (independent) is the DRINK CONTENT while the variable being measured (dependent) is the BLOOD PRESSURE. Hence, these variables serve as the template to ask a scientific question which goes thus:

HOW DOES THE DRINK CONTENT AFFECT AN INDIVIDUAL'S BLOOD PRESSURE?

This scientific question relates how the independent variable (drink) causes the dependent variable to respond (blood pressure).

3 0

2 years ago

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th

Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

$a=cs$ (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

$v^2=v_o^2+2a \Delta s$ (2)

vo: initial velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

$v^2=2(cs)\Delta s$

You do the constant c in the last equation, then you replace the values of v, s and Δs:

$c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44$

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3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

Covert 35000000 miles to meters and show how you got the answer?

Lina20 [59]

Answer:

56327040000 metres

Explanation:

1 mile =

1609.344 metres

35000000 miles = x meters

we represent x by the number of meters which the requested miles maps to

we cross multiply, so 1609.344×35000000 = 1 × x

x =56327040000 metres

8 0

2 years ago