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3 years ago

A sailfish swims 120 km/hr. How far will it travel in 8.0 minutes?

Physics

1 answer:

Rasek [7]3 years ago

5 0

Answer:

16km

Explanation:

First change the minutes into hours then multiply by the distance.

(8÷60)×120=16km

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A 1200-kg car moving at 15.6 m/s suddenly collides with a stationary car of mass 1500 kg. if the two vehicles lock together, wha

g100num [7]

Use conservation of momentum ;

m1u1 + m2u2 = m1v1 + m2v2

1200×15.6 + 0 = 2700v

v = 18720/2700

v = 6.933 or ~ 7 m/s

5 0

3 years ago

If toner particles in a laser printer have a negative charge, then what charge do you think the surface of the paper in the prin

Elina [12.6K]

Answer:

Explanation:

As it moves along, the paper is given a strong negative electrical charge by another corona wire. When the paper moves near the drum, its negative charge attracts the positively charged toner particles away from the drum.

3 0

2 years ago

A shopper pushes a 5.32 kg grocery cart

Juli2301 [7.4K]

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

$\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}$

$\text { Acceleration }=\frac{58.232}{5.32}$

$\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

7 0

3 years ago

Read 2 more answers

Why are “input” and “output” good words to use when discussing systems?

charle [14.2K]

Answer:

To determine how efficient that system is.

3 0

3 years ago

Read 2 more answers

What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant fig

Igoryamba

Answer:

$A=6.36\times 10^5\ m^2$

Explanation:

It is given that,

Radius of the circle, $r=4.5\times 10^4\ cm=4.5\times 10^2\ m$

The area of the circle is given by :

$A=\pi r^2$

$A=\pi (4.5\times 10^2\ m)^2$

$A=636172.51\ m^2$

$A=6.36\times 10^5\ m^2$

As there is no uncertainty given in the radius of the circle. So, the area of the circle is $6.36\times 10^5\ m^2$ . Hence, this is the required solution.

5 0

3 years ago