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2 years ago

J. Ram travels 50 m on a straight road in 30 s. Hari covers the same distance

Physics

1 answer:

ira [324]2 years ago

6 0

Explanation:

Explanation:

In the given question, there are two conditions.

Ram traveled 50 m in 30 seconds in a straight road.

Hari covers the same distance in the road; that means 50 m in 45 seconds

The formula for work w=\frac{d}{t}w=td

Ram's work w=\frac{50}{30}=1.66666w=3050=1.66666

Hari's work w=\frac{50}{45}=1.111w=4550=1.111

Ram is slower than Hari and the velocity of Ram is slower than Hari.

Ram has done less work than Hari.

Hence, Hari has done more work and he has more power than Ram.

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Apply the concept of density to explain why oil floats on water.

Sauron [17]

Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.

6 0

3 years ago

A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi

Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

$k = \frac{155}{0.10} N/m$

$k = 1550 N/m$

so now we have

$2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}$

$m = 8.11 kg$

so mass of the fish is 8.11 kg

4 0

3 years ago

Find the lengths of each of the following vectors

Irina18 [472]

Answer:

Explanation:

Generally, length of vector means the magnitude of the vector.

So, given a vector

R = a•i + b•j + c•k

Then, it magnitude can be caused using

|R|= √(a²+b²+c²)

So, applying this to each of the vector given.

(a) 2i + 4j + 3k

The length is

L = √(2²+4²+3²)

L = √(4+16+9)

L = √29

L = 5.385 unit

(b) 5i − 2j + k

Note that k means 1k

The length is

L = √(5²+(-2)²+1²)

Note that, -×- = +

L = √(25+4+1)

L = √30

L = 5.477 unit

Note that, since there is no component j implies that j component is 0

L = 2i + 0j - 1k

The length is

L = √(2²+0²+(-1)²)

L = √(4+0+1)

L = √5

L = 2.236 unit

(d) 5i

Same as above no is j-component and k-component

L = 5i + 0j + 0k

The length is

L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

L = 5 unit

(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

L = 3.742 unit

(f) i + j + k

The length is

L = √(1²+1²+1²)

L = √(1+1+1)

L = √3

L = 1.7321 unit

3 0

3 years ago

An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),

Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane $h=6\ miles$

When Airplane is $s=10\ miles$ away

Distance is changing at the rate of $\frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph$

From diagram we can write as

$h^2+x^2=s^2$

differentiate above equation w.r.t time

$2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}$

as altitude is not changing therefore $\frac{\mathrm{d} h}{\mathrm{d} t}=0$

$0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}$

at $s=10\ miles\ and\ h=6\ miles$

substitute the value we get $x=\sqrt{10^2-6^2}=8\ miles$

$8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290$

$\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph$

5 0

3 years ago

Two small nonconducting spheres have a total charge of 93.0 μC . Part A

Travka [436]

Answer:

charge on each

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

Q1Q2 = 1.49 × $10^{-9}$ C²

and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

5 0

3 years ago