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9 months ago

What causes an object to slowdown or speed up

Physics

1 answer:

ivanzaharov [21]9 months ago

7 0

Answer:

A force can speed up or slow down an object. A force can change the direction in which an object is moving. A bigger force on an object will produce a bigger change in the motion. A heavier object requires a larger force than a lighter object in order to undergo the same change in motion.

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When the Moon orbits Earth, what is the centripetal force?

nata0808 [166]

Answer:

Gravity is the centripetal force when the moon orbits the earth.

5 0

2 years ago

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

At 6: 00 am, a motorbike set off from town A to town B at a speed of 40km/h. At the same time, a car set off from town B to town

Keith_Richards [23]

Answer:

One would need to know how far apart the towns are:

T = SA / 40 time it takes for first cyclist to travel S1

T = SB / 60 time it takes for cyclist B to travel distance S2

SA + SB = S the distance between the towns

SB = 60 / 40 SA = 1.5 SA

SA + 1.5 SA = S

S = 2.5 SA where cyclist travels distance SA

The time will depend on the separation of the towns.

4 0

1 year ago

A laser source has a bandwidth of 30GHz (a) Calculate the coherence length of the source b) What is the time separation of secti

abruzzese [7]

Explanation:

It is given that,

Bandwidth of a laser source, $f=30\ GHz=30\times 10^9\ Hz$

(b) Let t is the time separation of sections of sections of the light wave that can still interfere. The time period is given by :

$T=\dfrac{1}{f}$

$T=\dfrac{1}{30\times 10^9}$

$T=3.33\times 10^{-11}\ s$

(a) Let h is the coherence length of the source. It is given by :

$l=c\times T$

c is the speed of light

$l=3\times 10^8\times 3.33\times 10^{-11}$

l = 0.0099 m

Hence, this is the required solution.

6 0

2 years ago

g Initially, the motorcycle travels along a straight road with a speed of 35 m/s (this is almost 80 mph). The maximum decelerati

astraxan [27]

Given:

Initial speed of the motorcycle (u) = 35 m/s

Final speed of the motorcycle (v) = 0 m/s (Complete Stop)

Maximum deceleration of the motorcycle (a) = -1.2 m/s²

Required Equation:

$\boxed{\bf{ v = u + at}}$

Answer:

By substituting values in the equation, we get:

$\rm \longrightarrow 0 = 35 + ( - 1.2)t \\ \\ \rm \longrightarrow 0 = 35 - 1.2t \\ \\ \rm \longrightarrow 35 - 1.2t = 0 \\ \\ \rm \longrightarrow 35- 35 - 1.2t = 0 - 35 \\ \\ \rm \longrightarrow - 1.2t = - 35 \\ \\ \rm \longrightarrow \dfrac{ - 1.2t}{ - 1.2} = \dfrac{ - 35}{ - 1.2} \\ \\ \rm \longrightarrow t = 29.167 \: s$

$\therefore$ Time taken by motorcycle to come to a complete stop (t) = 29.167 s

4 0

3 years ago