A.stationary <br> B. Accelerating <br> C. Decelerating <br> D. Moving at constant speed

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its

maxonik [38]

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

$KE=0.5mv^{2}$ where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

$KE=0.5*800*(20 m/s)^{2}=160,000$

Frictional force by vehicle pads is given by

$Fr=\frac {KE}{d}$ where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

$Fr=\frac {160000}{50}=3200 N$

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

7 0

3 years ago

(1) Expansion of concrete

Pachacha [2.7K]

Abcdefghijklmnopqurtuv

6 0

3 years ago

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

lesantik [10]

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0

3 years ago

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

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

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$