Question 12 (Mandatory) (5 points)

A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and

ycow [4]

Answer:

Acceleration a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

Explanation:

For the truck to accelerate without losing its load.

Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.

Fa ≤ F(friction)

But;

Fa = mass × acceleration

Fa = ma

ma ≤ F(friction)

a ≤ (F(friction))/m ......1

Given;

Fa = mass × acceleration

Fa = ma

mass m = 800 kg

F(friction) = 2400 N

Substituting the given values into equation 1;

a ≤ F(friction)/m

a ≤ 2400N/800kg

a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

4 0

3 years ago

Which of the following shows a balanced nuclear reaction?

miskamm [114]

Answer:

No. 2

Explanation:

6 0

3 years ago

Read 2 more answers

Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i

romanna [79]

Answer:

$v_f=0.825m/s$

Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

7 0

4 years ago

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

3 0

3 years ago

If car A passes car B, then car A must be

Tcecarenko [31]

Answer:

B. moving faster than car B, but not necessarily accelerating

Explanation:

Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.

4 0

3 years ago