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4 years ago

10

How many nanometers are 1 cm

1 answer:

frez [133]4 years ago

6 0

Answer:

The answer is 1e+7 NM.

Explanation:

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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

6 0

3 years ago

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

stiv31 [10]

Answer:

Magnetic field, $B=2\times 10^{-4}\ T$

Explanation:

Given that,

Velocity of electron, $v=5\times 10^7\ m/s$

It enters a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, $E=10^4\ V/m$

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, $F_m=qvB\ sin\theta$ .......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

$qvB\ sin\theta=qE$

$B=\dfrac{E}{v}$

$B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}$

B = 0.0002 T

or

$B=2\times 10^{-4}\ T$

Hence, this is the required solution.

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