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3 years ago

The rotational analogue of force is moment of force, also called torque.

Physics

1 answer:

Furkat [3]3 years ago

6 0

Answer:

the inertia would be the force equal mass times the acceleration

10 thieworhet

Explanation:

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Two football players with mass 75 kg and 100 kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively. if t

3241004551 [841]

heyaaaaa

Momentum before Pb = momentum after Pa

Pb = 75*6 - 100*8 = -350kgm/s = Pa = (75+100)V where V is the velocity of the combined mass of the two players after the collision.

Velocity has magnitude (speed) and direction. V = -350/175 = -2m/s

So the two players are moving at 2m/s in the direction the 100kg player was moving before the collision.

I arbitrarily chose the direction of the smaller player as positive so the opposite direction (of the larger player) had to be negative.

hope it helpssss!!!!!!

6 0

3 years ago

Question 2 (1 point)

mixer [17]

Answer:

The initial vertical velocity is zero, u = 0 m/s

Explanation:

Given;

height of the table, h = 0.55 m

horizontal distance traveled by the tennis, x = 0.12 m

Apply the following kinematic equation;

h = ut + ¹/₂gt²

where;

u is the initial vertical velocity = 0, since the tennis ball rolled off the edge of a table.

h = ¹/₂gt²

The time to fall from the vertical height is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(0.55)}{9.8} }\\\\t = 0.335 \ s$

The initial horizontal velocity of the tennis is given by;

x = vₓt

vₓ = x / t

vₓ = (0.12) / (0.335)

vₓ = 0.358 m/s

Therefore, the initial vertical velocity is zero, u = 0 m/s and initial horizontal velocity, vₓ is 0.358 m/s

3 0

3 years ago

Two blocks with mass M and 3M on a horizontal frictionless surface are pushed together and compress a spring of negligible mass

Naily [24]

Answer:

launching speed of the lighter block = -6 m/s

Explanation:

We are given;

Mass of light block; M

Mass of heavy block; 3M

Speed of launched block: v = 2m/s

We are told that the two blocks are sitting on the horizontal frictionless surface. Thus, we can say that no external force is being applied on the system and so, the momentum of the whole system is conserved accordingly to that condition.

We are also told that when the rope is cut with scissors, that the heavier block attains the speed of 2 m/s in the positive x-direction which is horizontal direction.

We know that formula for momentum is; M = mass x velocity.

Thus, the momentum of the heavier block is calculated as;

M_1 = 3M × 2

M_1 = 6M kg.m/s

Since no external force is applied on the object, the initial momentum will be zero.

Hence, to conserve the system, the momentum of the lighter block will be equal and opposite to the momentum of heavier block.

So, momentum of lighter block is;

M_2 = -6M kg.m/s

Since mass of lighter block is M and formula for momentum = mass x velocity.

Thus;

-6M = Mv

Where v is speed of lighter block.

So, v = -6M/M

v = -6 m/s

7 0

3 years ago

An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the

ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, $E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J$

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.

The magnetic field in north direction, $B_y=19911.5\ nT$

The magnetic field in west direction, $B_x=-3257.1\ nT$

The magnetic field in vertical direction, $B_z=48381.8 \ nT$

Magnetic field, $B=(-3257.1i+19911.5j+48381.8k)\ nT$

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

$E_k=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E_k}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

And, $i\times i=j\times j=k\times k=0$

$F=1.6\times 10^{-19}\times [1178 k-2864.20j]$

$|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}$

$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

$a=\dfrac{F}{m}$

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

$a=5.43\times 10^{14}\ m/s^2$

Hence, this is the required solution.

4 0

3 years ago

Part C: Quantitative Problems when vf is not 0

Alina [70]

Answer:

(a)

$\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s$

(b)

1120 N

Explanation:

Change in velocity, $\triangle v$ is given by subtracting the initial velocity from the final velocity and expressed as $\triangle v= v_f -v_i$

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

$\triangle v=0-8=-8\ m/s$