HELPPPPPURGENT PLEASE NB 35

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

8 0

3 years ago

What type of clouds are associated with low pressure?

Naddik [55]

Cumulus and cumulonimbus<span />

5 0

3 years ago

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

$V= \frac{Q}{C}$

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0

3 years ago

A race car starting from rest accelerates uniformly at a rare of 4.90 meters per second^2. What is the cars speed after it has t

Vesnalui [34]

From the law of Galileo Galilei :v²=v₀²+2ad we take the speed

v²=0+2*4.90*200=1960=>v=√1960=44.27 m/s

4 0

3 years ago

A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-do

sweet-ann [11.9K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{Work\:done=24\:J}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Constant \: force(F) = 8 \: N \\ \\ \tt: \implies Displacement(s) = 3 \: m \\ \\ \red{\underline{\bold{To \: Find : }}} \\ \tt: \implies Work \: Done(W.D) = ?$

• <u>According to given question</u> :

$\green{ \star} \tt \: \theta \: = 0 \degree \: \: \: \: (Angle \: between \: force \: and \: displacement) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: Done = FS \: cos \: \theta \\ \\ \tt: \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\ \\ \green{ \circ} \tt \: cos \: 0 \degree = 1 \\ \\ \tt: \implies Work \: Done =24 \times 1 \\ \\ \green{\tt: \implies Work \: Done =24 \: J}$

5 0

3 years ago