A car travels around an oval racetrack at constant speed. The car is accelerating:_______

Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh

ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0

2 years ago

Which term do modern psychologists prefer to use in place of short-term memory?

Vsevolod [243]

Correct answer: D). Working Memory

The short term memory is the part of the memory system in which the information is stored for 30 seconds. However, it can be increased by rehearsal. Short term memory is also called active memory or the working memory. The working memory is the part of the cognitive system that is responsible for storing information temporarily for processing. It is important for reasoning.

Hence, the correct answer would be option D.

7 0

2 years ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

Ad libitum [116K]

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

7 0

2 years ago

A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material

worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

$r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m$ is the radius

$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

$d=\frac{m}{V}$

where m = 1 kg is the mass. Substituting, we find

$d=\frac{1 kg}{7.1\cdot 10^{-5} m^3}=14,160 kg/m^3$

3 0

2 years ago

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ fro

netineya [11]

Refer to the diagram shown below.

v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ = the angle that the string makes with the vertical
m = Bob's mass (mg = the weight)
F = centripetal force
l = the length of the string

From geometry,
r = l sin θ

The centripetal acceleration is
$a= \frac{v^{2}}{r} = \frac{v^{2}}{l \,sin \theta}$
The centripetal force is
$F = \frac{m v^{2}}{l \, sin \theta}$

For vertical force balance,
T cosθ = mg (1)
For horizontal force balance,
$Tsin \theta = F = \frac{m v^{2}}{l sin \theta}$ (2)
Divide (2) by (1).
$tan \theta = \frac{v^{2}}{gl sin \theta} \\\\ v^{2} = gl\, sin \theta \,tan \theta \\\\ v= \sqrt{gl \, sin \theta \, tan \theta}$

Answer: $v = \sqrt{gl \, sin \theta \, tan \theta}$

5 0

1 year ago

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A car travels around an oval racetrack at constant speed. The car is accelerating:________.