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umka2103 [35]
3 years ago
13

An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant spe

ed
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

Increasing speed.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

In this scenario, an object moves with a positive acceleration. Thus, the object is moving with an increasing speed and as such it has acceleration in the same direction as its velocity with respect to time.

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Which arrow best represents the path of an object with projectile motion?
Fittoniya [83]
C because it's non linear
7 0
3 years ago
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What is the position of the car?
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An object slides down a very smooth ramp, with negligible friction. It slides with constant acceleration a under the action of t
sveta [45]

Answer:

In statics condition and if is a body is over plane surface N = mg

In all cases which a body is on a ramp force N will be smaller than mg, and depending on the level of the ramp (bigger the ramp smaller force N)

Explanation:

See annex (Figure 1)

We can see with the help of force diagram the relation between  N force  and mg

Lets look first the body on the ramp assuming there is not movement

The  sum of forces alog y axis = 0

mg* cos∠ BAC = N (normal force)

So N is always a fuction of mg and of the cosine of the inclination angle, and the cosine of angle varies from value 1 when the angle is 0 (look the case where the block is on the surface without inclination N = mg ).

And as the inclination increase the value of cosine will become smaller and so will N force which is directly proportional of cosine of the inclination angle

Therefore we can say that force N is always minor than mg unless in extreme case where the block is over the surface in which case N takes its maximum value N = mg

3 0
4 years ago
A soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill
ozzi

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

                             S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

                            h = S²g/2Vx²                     (Since Vy = 0)

Substituting the values

                             h = 42² x 9.8/ (2 x 15²)

                                = 38.42 m

Hence, the height of the hill is, h = 38.42 m

4 0
3 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
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