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3 years ago

The value of a car after 5 years is a quantitative data. True or False

Physics

2 answers:

NeX [460]3 years ago

7 0

Answer:

true

Explanation:

gladu [14]3 years ago

7 0

Answer:

True or False: True

Explanation:

Quantitative Deals With Numbers(5)

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It takes a ball 30 seconds to travel from point A to point B. What is the ball’s speed?

AlladinOne [14]

Direction from A to B divided by time(30s)

3 0

3 years ago

What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a

Elena L [17]

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

6 0

3 years ago

Calculate the molecular weight of Aluminium hydroxide

Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

5 0

3 years ago

Harry Potter is chasing his nemesis Draco Malfoy during a quidditch match. Initially, Harry is 35m behind Draco, who has just sp

Sophie [7]

Answer:

the acceleration of harry is equal to 66.126 m/s²

Explanation:

given,

harry is 35 m behind Draco

speed of Draco = 40 m/s

original speed of harry = 50 m/s

acceleration = ?

time taken by the Draco

t = $\dfrac{r}{u} =\dfrac{75}{40}$

t = 1.875 s

distance covered by Harry

d = 35 + 175 = 210 m

to calculate the acceleration of harry

$s = u t+ \dfrac{1}{2}at^2$

$210 = 50\times 1.875+ \dfrac{1}{2}\times a\times 1.875^2$

a × 3.516 × 0.5 = 116.25

a = 66.126 m/s²

hence, the acceleration of harry is equal to 66.126 m/s²

3 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$