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3 years ago

13

Which requires more energy: lifting a 50 kg sack vertically 2 meters or lifting a 25 kg sack vertically 4 meters?

2 answers:

gtnhenbr [62]3 years ago

5 0

Answer: C) both require the same amount of work

Explanation:

If you think about 25 is half of 50 and 2 is half of 4 and 50 times 2 =100 and

25 times 4 =100

elena-s [515]3 years ago

3 0

Answer:

c. Both require the same amount of work

Explanation:

GPE = mgh

50 * 10 * 2 = 1000 J

25 * 10 * 4 = 1000 J

Therefore, both are same and require the same amount of work.

Thenk and mark me brainliest :)

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A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t

vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be $m$ kg.

Given:

Initial velocity of the car is, $u=25\ m/s$

As the car stops, final velocity of the car is, $v=0\ m/s$

Change in momentum is, $\Delta p=39400$

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

$\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg$

Therefore, the mass of the car is 1576 kg.

4 0

3 years ago

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

What is the approximate efficiency of the engine?

Elodia [21]

Answer:91% I think

Explanation:

8 0

3 years ago

Read 2 more answers

Critical angle of glass is 42 .what does it mean?

elena55 [62]

Answer:

i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!

Explanation:

HOPE IT HELPS!!

6 0

3 years ago

A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

2 years ago