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Tanzania [10]
3 years ago
7

which of the following is not a projectile? a.a satellite b.a throw ball c.a ball on the ground d.a soaring arrow

Physics
2 answers:
Delvig [45]3 years ago
7 0
C a ball on the ground
kirza4 [7]3 years ago
4 0
A projectile is known as an object that is thrown in the air or shot in the air....for example a rock that is thrown up from a tall building or a missile that is shot from a rocket launcher. A projectile is usually also considered an object that is thrown up and then comes back down.....using the examples above....if a rock is thrown up eventually it is going to have to come down....same thing for the missile.....

I believe the answer is B.

hope this helps :)


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How many moles are in 95 g of Na?
nexus9112 [7]
1.8 mol of Na. hope this helps
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3 years ago
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initially, a bowl holds 15 m^3 of water. an object is dropped into the bowl and the new volume of the water is 25 m^3. what is t
marusya05 [52]

Explanation:

The new volume of water = 25 ml

The old volume of water = 15 ml

The difference = 25 - 15 but what are the units?

Since the question asks for force, the units must start out as 10 mL

In water 1 mL has a mass of 1 gram, so the answer is 10 grams.

Grams are units of mass, not weight. You should convert this into newtons.

10 grams = 1/1000 = 0.01 kg

1 kg has a weight of 9.81 Newtons

0.01 kg has a weight 0.081 Newtons

If you have never seen a Newton before, then the answer is 10 grams

3 0
3 years ago
A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co
exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0
3 years ago
How to solve arctan[tan(7pi /4)] ...?
Svet_ta [14]
Well, 
arctan is a bijection from R into (-pi/2 , pi/2)*and 
pi is a period of tangent function: 
so 
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4) 
we finally get : 

<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4 
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
7 0
3 years ago
sla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have different
sweet [91]
Because acceleration depends not only on the change in velocity.
It also depends on the time during which the change occurs.
The formula is

Acceleration = (change in velocity) divided by (time for the change) .

Maybe Sla changed his velocity in 3 seconds, but Hazel
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so the quotients of (change/time) were different. 
8 0
3 years ago
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