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3 years ago

13

URGENT

1 answer:

Advocard [28]3 years ago

5 0

Answer:

$give \\ mass(m) = 1350kg \\ acceleration(a) = 1ms {}^{ - 2} \ \\ sln\ \\ from \: our \: formula \\ \: f = ma \\1350kg \times 1ms {}^{ - 2} \\ f = 1350newton$

the force you applied to your car =1350N

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A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. T

krok68 [10]

Answer:

the correct answer is B

Explanation:

Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved

instantly starts. Before the crash

p₀ = M v +0

final instant. After the crash

m_f = (M + M) v_f

the moment is preserved

p₀ = p_f

M v = 2 M v_f

v_f = v / 2

let's look for kinetic energy

before the crash

K₀ = ½ M v²

after the crash

K_f = ½ 2M (v_f)²

K_f = ½ 2M (v/2)²

K_f = (½ M v²) ½

K_f = K₀ / 2

therefore the correct answer is B

4 0

3 years ago

If a proton and an electron are released when they are 4.00 x 10^-10 m apart, find the initial acceleration of each of them.

SVEN [57.7K]

Answer:

For proton

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=1.5\times 10^{21}\ m/s^2$

Explanation:

We know that

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The mass of proton

$m=1.67\times 10^{-27}\ kg$

Charge on electron and proton

q₁=q₂=q

$q=1.6\times 10^{-19}\ C$

Electrostatics force

$F=K\dfrac{q_1q_2}{r^2}$

Now by putting the values

$F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}$

$F=1.44\times 10^{-9}\ N$

For proton

F = m a

a =F/m

$a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2$

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2$

$a=1.5\times 10^{21}\ m/s^2$

5 0

3 years ago

Read 2 more answers

11. Newton's Third Law of Motion relates to action and reaction. Which of the following scenarios accurately names the

hoa [83]

Answer:

b

Explanation:

4 0

3 years ago

A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Tresset [83]

Answer:

$\theta_1 = 0.400^o$

$\theta_2 =0.378^o$

Explanation:

From the question we are told that

The number of slits per cm is k = $161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m$

The order of the maxima is n = 1

The wavelength are $\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m$

The spacing between the slit is mathematically represented as

$d = \frac{ 0.01}{k}$

=> $d = \frac{ 0.01}{161}$

=> $d = 6.211 *10^{-5} \ m$

Generally the condition for constructive interference is

$n\lambda = d \ sin \theta$

At $\lambda_1$

$\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_1 = 0.400^o$

At $\lambda_2$

$\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_2 =0.378^o$

4 0

3 years ago

Find the moment of force of 60 Newton about an axis of rotation at distance 20 cm from the force

777dan777 [17]

Answer:

10 N.m

Explanation:

Force * Perpendicular distance.

= 20 N * 0.5 m

= 10 N.m

8 0

3 years ago