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3 years ago

a car is traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is the acceleration?

Physics

2 answers:

torisob [31]3 years ago

5 0

I can’t see if anyone have gave you the answer yet

ad-work [718]3 years ago

4 0

-0.5 m/s^2

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please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

6 0

3 years ago

Methods to determine the specific heat capacity of a substance

Gre4nikov [31]

The heat capacity and the specific heat

5 0

3 years ago

Read 2 more answers

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

2 years ago

The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

Ainat [17]

When t=2, the ball has fallen   d(2) = 16 (2²) = 64 feet .

When t=5, the ball has fallen   d(5) = 16 (5²) = 400 feet .

Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.

Time period between t=2 until t=5 is (5 - 2) = 3 seconds.

Average speed of the ball from t=2 until t=5 is

   (distance covered) / (time to cover the distance)

   =    336 feet    /    3 seconds = 112 feet per second.

That's what choice-C says.

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3 years ago

If a bean of mass 2.0g jumps 1.0cm from your hand into the air, how much potential energy has it gained in reaching its highest

kirill [66]

$PE = mg\Delta h = 0.002 \, kg \cdot 9.8 \, m/s^2 \cdot 0.01 \, m = ~2 \cdot 10^{-4} \, J$

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3 years ago