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3 years ago

What are 2 ways acceleration can be changed?

1 answer:

3 0

2 Ways Acceleration can be changed are:

i) Increasing or decreasing the velocity of a moving object.

ii) Change of direction of the moving object.

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two friends are talking to another person across the building the other person can hear them but not see them why is this?

Rufina [12.5K]

Answer:

Hey

Say that there was no light on in the building that they were ch,at,tin,g in, then they could hear each other but not see each other.

4 0

3 years ago

9. A car driver brakes gently. Her car slows down front --

sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0

3 years ago

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

$P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,$

where $h_1$ and $h_2$ are the height of the column of oil and the unkown liquid, respectively. Writing for $\gamma_{unk}$ , we have

$\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.$

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0

2 years ago

Select the correct answer from each drop-down menu.

kirill115 [55]

Answer:

Explanation:

I think but I am sure

8 0

3 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

2 years ago