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Effectus [21]
3 years ago
12

Which of two factors influence the weight of an object due to gravitational pull?

Physics
1 answer:
Maslowich3 years ago
5 0
Where are the factors ... to this question

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Ariana is accelerating her car at a rate of 4.6 m/s2 for 10 seconds. Her starting velocity was 0 m/s.
Oksanka [162]

answer is 46 m/s

hope it help you

3 0
3 years ago
In the diagram, ∠D is b
Degger [83]

Answer:

The answer is B

Explanation:

3 0
3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the
arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

s_1=ut_1+\frac{1}{2}a_1 t_1^2

where:

u = 0 is the initial velocity (the car starts from rest)

t_1 = 20 s is the time elapsed in the 1st part

a_1=1.62 m/s^2 is the acceleration of the car in the 1st part

s_1 is the displacement of the car in the 1st part

Solving for s_1,

s_1=0+\frac{1}{2}(1.62)(20)^2=324 m

We can also find the velocity of the car after these 20 seconds using the equation:

v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

s_2 = (\frac{v_1 + v_2}{2})t_2

where:

v_1=32.4 m/s is the initial velocity at the beginning of the 2nd phase

v_2=0 is the final velocity (the car comes to a stop)

t_2=5 s is the time elapsed in the 2nd phase

Substituting,

s_2=\frac{32.4+0}{2}(5)=81 m

So, the total displacement of the car is

s=s_1+s_2=324+81=405 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
What is your least favorite candy? Explain
aev [14]

Answer:

My least favorite is whoppers.....Trust me i love chocolate, but not when it taste like chalk......

7 0
3 years ago
Read 2 more answers
If the average velocity during the athlete's walk back
goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

4 0
2 years ago
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