A car traveling at 20 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is it’s acceleration?

James Bond is trying to escape his enemy on a speedboat but

SVEN [57.7K]

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Putting all the values, we get :

$0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s$

So, Bond's boat is moving with a speed of 100 m/s after the collision.

3 0

3 years ago

Solve the inequality. x/3 is greater than or equal to - 6. a. x ≥ –9 b. x ≥ 9 c. x ≥ –18 d. x ≤ –18

NISA [10]

To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS.

~ x greater than or equal to -18

(C)

6 0

3 years ago

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

b)

We know that

F = m a = E q ------------1

Mass of electron ,m

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

Charge on electron

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

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0

3 years ago

Read 2 more answers

5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw

abruzzese [7]

For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle is mathematically given as

$\theta=n(\lambda/d)$

Therefore

$\theta 1/ \theta 2=n1(\lambda/d)/ n2(\lambda/d)$

In conclusion

theta/7=16/5

theta=10*7/5

theta=25.3