All categories

3 years ago

How are the variables speed and velocity different? how are they similar

2 answers:

7 0

Speed, (scalar quantity), is the rate at which an object covers distance. The average speed is the distance (scalar quantity) per time ratio. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (vector quantity) per time ratio.

victus00 [196]3 years ago

3 0

Answer:

Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. ... Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Explanation:

You might be interested in

A 100 kg object and a 10 kg object are dropped simultaneously in a vacuum. Which of the following statements is true?

ella [17]

In a vacuum, all objects fall at the same rate. Meaning that the 100 kg ball will fall at the same speed as the 10 kg ball. Assuming that both objects share the same starting acceleration, they will keep that acceleration until the fall is stopped.
In other words, your answer is the first one, Both objects will accelerate at 9.8 m/s

7 0

3 years ago

Read 2 more answers

If a car has a momentum of 1000kgm/s and<br> velocity of 500m/s, what is its mass

Dvinal [7]

Answer:

hey u apply p=mv and 2 are given then calculate thirds value it's a simple do it

7 0

3 years ago

What is a normal force?

Flauer [41]

Is the component perpendicular to the surface on contact of the contact force <span />

3 0

4 years ago

What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10

raketka [301]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

$Y = \frac{FL}{A\Delta L}$

$20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}$

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

7 0

4 years ago

During her high bar routine from question 2, Gabby Douglas slipped and falls from the high bar, landing on a 10 cm thick gymnast

tankabanditka [31]

a) 122.5 J

b) -122.5 J

c) -1884.6 N

d) -3769.2 N

e) $-753.8 m/s^2$

f) $a=-76.9 g$

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion.

Mathematically, it is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

Here, we want to find the kinetic energy of the head just before hitting the mat.

At that instant, the speed is:

v = 7 m/s

The mass of the head is:

m = 5 kg

So, the kinetic energy is

$K=\frac{1}{2}(5)(7)^2=122.5 J$

According to the work-energy theorem, the work done by a force on an object is equal to the change in kinetic energy of the object:

$W=K_f - K_i$

where

W is the work done

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

In this problem:

$K_i=122.5 J$ is the kinetic energy of the head just before hitting the mat

$K_f=0 J$ is the final kinetic energy (since the head comes to a stop)

So, the work done by the mat is:

$W=0-122.5 = -122.5 J$

The work is negative because the force exerted by the mat is opposite to the direction of motion of the head.

The work exerted by a force on an object is given by

$W=Fd$

where

F is the force applied

d is the displacement of the object

W is the work done

In this problem:

W = -122.5 J is the work done by the mat on the head

d = 6.5 cm = 0.065 m is the displacement of the head (since it deflects the mat by this amount)

So, the average force exerted by the mat on the head is:

$F=\frac{W}{d}=\frac{-122.5}{0.065}=-1884.6 N$

(the negative sign indicates that the force is in direction opposite to the motion of the head)

The force calculated in part d) represents the average force exerted by the mat on the head:

$F_{avg}=-1884.6 N$

We can assume that as the head first hits the mat, the initial force is zero, then increases at a constant rate up to a peak value of $F_{peak}$ , then it decreases again until the head stops.