Sal sprinted 40 m to the right in 5.5s what is his average velocity

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

2 years ago

A ruler of length 0.30m is pivoted at its centre. Equal and opposite forces of magnitude 2.0N are applied to the ends of the rul

kow [346]

Answer:

0.3858 Nm

Explanation:

The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

$T = \vec{F} \cdot \vec{s} = Fscos(50^0) = 2 * 0.3 * 0.643 = 0.3858 Nm$

6 0

3 years ago

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

$y(t)=2.80t+0.61t^3$

The velocity of the crate vs time is given by the derivative of the position, so it is:

$v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2$

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

$a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t$ [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

$F(t)=ma(t)$

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

$a(4.10)=3.66\cdot 4.10 =15.0 m/s^2$

And therefore, the force on the crate is:

$F=ma=(5.00)(15.0)=75 N$

7 0

3 years ago

What would be the final size of a balloon, at -18*C, if the initial volume is 2.0 L and the initial temperature is 20*C?

Anuta_ua [19.1K]

Answer:

1.7 L

Explanation:

PV = nRT

If P, n, and R are constant:

V₁ / T₁ = V₂ / T₂

(2.0 L) / (293.15 K) = V / (255.15 K)

V = 1.7 L

7 0

3 years ago

Water has a density of 1000 kg/m3. A piece of rubber has a density

almond37 [142]

Explanation:

if the rubber was put in a pool of water then the rubber sinks in water because the density of rubber (1024 kg/m³) Is more than that of water (100 kg/m³) . As we know if the density is greater than water then the substance sinks in water.

Hope it will help :)

3 0

2 years ago