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Murrr4er [49]
2 years ago
11

Hey guys can you help me solve this problem "how long will it take a car travelling 30m/s to come to stop ifs its acceleration i

s -3 m/s2".
Physics
1 answer:
salantis [7]2 years ago
7 0

Answer:

10 seconds.

Explanation:

We can use a kinematic equation where we know the final velocity, initial velocity, acceleration, and need to determine the time <em>t: </em>

<em />\displaystyle v_f = v_i + at<em />

<em />

The initial velocit is 30 m/s, the final velocity is 0 m/s (as we stopped), and the acceleration is -3 m/s².

Substitute and solve for <em>t: </em>

<em />\displaystyle \begin{aligned} (0\text{ m/s}) & = (30 \text{ m/s}) + (-3 \text{ m/s$^2$}) t \\ \\  t & = \frac{-30\text{ m/s}}{-3 \text{ m/s$^2$}} \\ \\ & = 10 \text{ s} \end{aligned}<em />

<em />

Hence, it will take the car 10 seconds to come to a stop.

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If an object is projected upward from ground level with an initial velocity of 144144 ft per​ sec, then its height in feet after
Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

v_0 = 144 ft/s

The equation that describes its height at time t is

s(t) = -16t^2 + 144 t (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

v(t) = s'(t) = -32t +144 (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

0=-32t+144\\t=\frac{144}{32}=4.5 s

And by substituting this value into eq.(1), we also find the maximum height:

s(t) = -16(4.5)^2+144(4.5)=324 ft

3 0
3 years ago
Maggie's truck decelerates from 18.00 m/s to rest in 3.30 s. If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculat
Free_Kalibri [48]

Answer:

look at my Explanation

Explanation:

If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculate the magnitude of the net (unbalanced) force that can cause the acceleration.

8 0
2 years ago
Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t
Viefleur [7K]
It’s is D bc it is and idk you should do it
6 0
3 years ago
PLS HELP DUE AT 10 PM
Rina8888 [55]
I believe it is C hope i helped!
6 0
3 years ago
Read 2 more answers
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
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