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Karolina [17]
2 years ago
7

Which of the following units could be used for the average speed of an object? A. kilometers per hour (km/hr) B. miles per hour

(mi/hr, mph) C. meters per second (m/s) D. all of these
Physics
1 answer:
Shkiper50 [21]2 years ago
5 0

Answer : The correct option is, (D) all of these

Explanation:

Average speed : It is defined as the ratio total distance to total time.  It is a scalar quantity that means it is defined by the magnitude only.

Formula used :

\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}

The common units of average speed are miles per hour (mph), kilometers per hour (kph), meters per second (m/s), or feet per second (ft/s).

The S.I unit of average speed is meter/second (m/s).

Hence, all of the units could be used for the average speed of an object.

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John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0
3 years ago
What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2
jasenka [17]

Answer:

5 kg

Explanation:

Acceleration = 6 m/s^2

Force = 30 N

Force = mass * acceleration

mass = force / acceleration

mass = 30 / 6

mass = 5 kg

4 0
2 years ago
Can somebody please explain how a compass can be used to find North?
Ira Lisetskai [31]

if the pointy thingy in your compass is pointing north, that means it's being (pulled toward) something near Earth's north pole

5 0
2 years ago
Which is not a common extrusive igneous rock
KonstantinChe [14]
Granite is the answer i think
8 0
2 years ago
An object is 15 cm in front of a diverging lens with a
Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0
3 years ago
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