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3 years ago

A girl is walking 2km north them east then south then west whar is her displacement

1 answer:

6 0

You haven't told us the lengths of the last 3 pieces.

IF all 4 pieces are the same length, then she ends up
in exactly the same place where she started, and her
displacement for the complete square walk is zero.

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The water cycle was described in the Scriptures long before modern scientists understood it.

s344n2d4d5 [400]

Answer:

False

Explanation:

Low humidity is typically associated with cold temperatures since cold air holds less water compared to warm air. While it is possible for indoor humidity to be as low as 10%, most people generally find humidity levels of about 40 to 45 percent comfortable.

4 0

2 years ago

Read 2 more answers

_________ is the increase in density of a medium due to wave travel.

Alekssandra [29.7K]

Answer:

<h2>compression</h2>

Explanation:

Such type of the longitudinal waves that causes compression during the travelling through the medium is called as compression. Generally, displacement of the medium takes place in this wave travelling conditions. There are many waves that are considered as longitudinal waves such as sound waves, seismic p-waves and some other. Waves have different types of properties that are changed due to different medium.

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3 years ago

Read 2 more answers

A person travelled in the -x direction at a constant speed of 2.5 m/s. The person's final position is -10.0 m, and the initial p

Nostrana [21]

Answer:

Explanation:

We have given the final position = -10 m on x axis

And the initial position =10 m

So total distance = 10-(-10)=20 m

The speed is given as 2.5 m/sec

We have tof ind the time required by the person to travel

Time is given by $t=\frac{distance}{speed}=\frac{20}{2.5}=8 sec$

So the option (c) is correct option

4 0

3 years ago

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

The height is $H = 6.74 \ m$

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$

$-1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2$

=> $-4.9t^2 + 9.6418t + 1 = 0$

using quadratic formula to solve the equation we have

$t = 2.07 \ s$

Generally distance covered along the horizontal is

$D = v cos (40) * 2.07$

=> $D = 15 cos (40) * 2.07$

=> $D = 23.74 \ m$

7 0

3 years ago

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Vedmedyk [2.9K]

Answer:

(a) V (A) = 0.7 m/s,

(b) V (A) = 0.7 m/s,

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a) law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0

3 years ago