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MissTica
3 years ago
14

Over a span of 6.0 seconds, a car changes its speed from 89 km/h to 37 km/h. What is its average acceleration in meters per seco

nd squared (m/s2)?
Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer: Average acceleration is -2.407 \frac{m}{s^{2}}

 

Explanation:

Average acceleration is equal to change in velocity divided by time taken

therefore, a=\frac{Vf-Vi}{t}

=> a=\frac{37-89}{6}\times \frac{5}{18}\frac{m}{s^{2}} = -2.407 \frac{m}{s^{2}}

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If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
What must be true about a surface in order for diffuse reflection to occur?
balu736 [363]

Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

4 0
3 years ago
If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t
Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to  Biot-Savart law, A magnetic field generated by a current  carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

 μ₀ =permeability of free space  4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 =  μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50  X 10 ^-3 X 2π  X  X 12.8 X 10^-2/  4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0
3 years ago
The percentage of incident light radiation reflected back to space is termed albedo. True or false?
liubo4ka [24]

Answer:True

Explanation:

Albedo is a unit-less, non-dimensional quantity that shows how well a surface reflects solar energy. The value of albedo can vary from 0 to 1, 0 being the black and 1 refers to a white surface. Zero means Surface is a perfect absorber i.e. it absorbs all the incoming rays incidents on it. Albedo 1 means the surface is a perfect reflector.

Albedo usually applies for visible light, even though it may involve some of the infrared regions of the electromagnetic spectrum. The average albedo associated with earth surface is 30%

8 0
3 years ago
An astronaut on a strange new planet having no atmosphere finds that she can jump up to a maximum height of 27 m when her initia
vekshin1
The magnitude is 4.5
7 0
3 years ago
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