The majority of pelagic sediments are of which type?

What is the momentum of a ten ton truck which is sitting still?

ANTONII [103]

The momentum of the truck is given by:

p = mv

p = momentum, m = mass, v = velocity

There isn't too much work to do here; the truck is at rest, therefore its velocity is 0. This means the product of m and v is 0, giving the momentum as 0.

7 0

3 years ago

A wire perpendicular to the screen carries a current into the screen (away from you). What is the direction of the magnetic fiel

Llana [10]

Answer:

The direction is downwards. This answer to this question is found from the applicatikn of flemings right hand rule. This rule uses the thumb, index finger and the middle finger in the direction of current force and the magnetic field.

Explanation:

4 0

3 years ago

car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s

kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

$-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})$

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

$-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]$

4 0

2 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can

Vera_Pavlovna [14]

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :

$a=\omega^2R$

Since, $\omega=\dfrac{2\pi }{T}R$

$a=(\dfrac{2\pi}{T})^2R$

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

$a'=(\dfrac{2\pi}{T'})^2R'$

$a'=(\dfrac{2\pi}{(2T)})^2(6R)$

$a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R$

$a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R$

Hence, this is the required solution.

4 0

2 years ago

A small sphere is hung by a string from the ceiling of a van. When the van is stationary, the sphere hangs vertically. However,

olganol [36]

Answer with Explanation:

We are given that

String makes an angle w.r.t vertical= $\theta=$

a.We have to derive an expression for the magnitude of the acceleration of the van in terms of the angle $\theta$ and magnitude g of the acceleration due to gravity.