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Delicious77 [7]

3 years ago

14

Imamu’s car is travelling twice as fast as Mosi’s car. Mosi travels 200 km in 2 hours. How long does it take Imamu to travel 50

km?

1 answer:

soldi70 [24.7K]3 years ago

8 0

Answer:

0.25hr

Explanation:

Given parameters:

Distance = 200km

Time = 2hrs

Now, let us find the rate which is the speed;

Speed = $\frac{distance}{time}$

Insert parameters:

Speed = $\frac{200}{2}$ = 100km/hr

Since the speed of Imamu = 2 x speed of Mosi = 200km/hr

Time taken = $\frac{distance}{speed}$ = $\frac{50}{200}$ = 0.25hr

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A proton is observed to have an instantaneous acceleration 11*10^11. what is the magnitude of e of the electric field at the pro

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The magnitude of the electric field at the proton's location is 10,437.5 N/C.

<h3>What the magnitude of the electric field?</h3>

The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.

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2 years ago

If the rest energies of a positive sigma particle and a lambda particle are 1189.4 and 1115.7 Mev respectively, what is the diff

Lunna [17]

Answer:

1.31022×10⁻³⁴ kg

Explanation:

Difference in energies = 1189.4-1115.7 MeV = 73.7 MeV

Convert this energy to Joules

1 MeV = 1.6×10⁻¹⁹ J

73.7 MeV = 73.7×1.6×10⁻¹⁹ J = 117.92×10⁻¹⁹ J

c = Speed of light = 3×10⁸ m/s

From Einstein's equation

E = mc²

$m=\frac{E}{c^2}\\\Rightarrow m=\frac{117.92\times 10^{-19}}{(3\times 10^8)^2}\\\Rightarrow m=1.31022\times 10^{-34}\ kg$

∴ Difference in their masses is 1.31022×10⁻³⁴ kg

8 0

3 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

1)

<u>Speed of blue car after 2.6 seconds of starting the motion:</u>

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

<u>Speed of blue car after 7.2 seconds of starting the motion:</u>

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

<u>Distance travelled by the blue car before application of brakes:</u>

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

<em>Now the distance travelled during the accelerated motion:</em>

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

<em>Now the distance travelled at uniform motion:</em>

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

<u>Acceleration of the blue car once the brakes are applied</u>

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

<u>The total time for which the blue car moves:</u>

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

<em>Now the time taken to stop the blue car after application of brakes:</em>

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

<u>For the acceleration of the yellow car:</u>

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

<em>Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.</em>

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

3 years ago

A ball is thrown upwards from the edge of a cliff. The horizontal velocity and vertical velocity are both 20m/s the distance fro

german

Answer:

20m

6.9s

Explanation:

The vertical velocity of the ball is 20m/s. We can calculate the kinetic energy which gets transferred to potential energy once it gets to the top.

$E_k = E_p$

$0.5mv^2 = mgh$

$h = \frac{0.5v^2}{g}$

we can subtitute v = 20m/s and g = 10m/s2

$h = \frac{0.5*20^2}{10} = 20 m$

So the ball could go 20m high from the child hand, or 120m fro the bottom of the cliff.

The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2

t = v / g = 20 / 10 = 2s

Then the ball will start accelerating down ward with a constant acceleration of g = 10m/s. In order to cover distance d of 120m from the top to the bottom of the cliff

$d = \frac{gt_2^2}{2}$

$t^2 = \frac{2d}{g} = \frac{2*120}{10} = 24$

$t = \sqrt{24} = 4.9s$

So the total time it takes is 4.9 + 2 = 6.9s

3 0

3 years ago

An escalator carries you from one level to the next in an airport terminal. The upper level is 4.7 m above the lower level, and

Assoli18 [71]

Answer:

2812.66685 J

-2812.66685 J

Explanation:

m = Mass of person = 61 kg

L = Length of escalator = 7 m

h = Height of escalator = 4.7 m

The angle between the normal force and the distance traveled by the person would be

$\theta=cos^{-1}\frac{h}{L}\\\Rightarrow \theta=cos^{-1}\frac{4.7}{7}\\\Rightarrow \theta=47.82^{\circ}$

Work done is given by

$W=mgLcos\theta\\\Rightarrow W=61\times 9.81\times 7\times cos47.82\\\Rightarrow W=2812.66685\ J$

The work done by the escalator on the when the person rides it up is 2812.66685 J

While going down the angle will be

$\theta=180-47.82\\\Rightarrow \theta=132.18^{\circ}$

$W=mgLcos\theta\\\Rightarrow W=61\times 9.81\times 7\times cos132.18\\\Rightarrow W=-2812.66685\ J$

The work done by the escalator on the when the person rides it down is -2812.66685 J

3 0

3 years ago