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2 years ago

A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?

Physics

1 answer:

Delvig [45]2 years ago

8 0

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

Explanation:

a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

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Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

7 0

3 years ago

A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag

Kobotan [32]

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5 0

3 years ago

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle

tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

$-h= vsinA\times t-gt^2/2$

putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = $-H =v sinA t - gt^2/2$

⇒ $4.9t^2 - 8.9\times0.8t - 100 = 0$

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0

3 years ago

calculate your power in watts if you do 500 N*m of work picking up a dumbbell,and takes 5 s to perform that task?

Slav-nsk [51]

Answer:

100watts

Explanation:

Given parameters:

Workdone = 500Nm

Time taken = 5s

Unknown:

Power in watts = ?

Solution:

Power is the rate at which work is done;

Power = $\frac{workdone }{time}$

Input the parameters and solve;

Power = $\frac{500}{5}$ = 100watts

6 0

3 years ago

Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. Straight upward incline with a speed of 2.95

Verdich [7]

Answer:

The length of the incline is 3.504 meters.

Explanation:

Let suppose that Julietta's ball decelerates uniformly, then we determine the length of the incline is determined by the following equation of motion:

$\Delta s = v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (Eq. 1)