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PilotLPTM [1.2K]
3 years ago
6

What is the acceleration if a car slows down from 30 m/s to 15 m/s in 4 seconds?

Physics
2 answers:
Tamiku [17]3 years ago
4 0

Answer:

11.25

Explanation:

acceleration =v-u/t

=-30-15/4

=-45/4

=11.25

hope it helps

pantera1 [17]3 years ago
3 0

Answer:

\boxed{\sf Acceleration \ (a) = -3.75 \ m/s^2}

Given:

Initial velocity (u) = 30 m/s

Final velocity (v) = 15 m/s

Time taken (t) = 4 seconds

To Find:

Acceleration (a) of the car

Explanation:

From equation of motion we have:

\boxed{ \bold{v = u + at}}

By substituting value of v, u & t in the equation we get:

\sf \implies 15 = 30 + 4a \\  \\  \sf \implies 4a + 30 = 15 \\  \\  \sf \implies 4a = 15 - 30 \\  \\  \sf \implies 4a =  - 15 \\  \\  \sf \implies a =  -  \frac{15}{4}  \\  \\  \sf \implies a =  - 3.75 \: m {s}^{ - 2}

\therefore

Acceleration (a) of the car = -3.75 m/s²

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Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

k = \frac{F}{\Delta x}\\

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\

For a load of 30 N, applying Hooke's Law:

\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\

Hence, the correct option is:

<u>B. 6 cm</u>

7 0
2 years ago
The current in two identical light bulbs connected in series is 0. 25 A. The voltage across both bulbs is 110 V. The resistance
konstantin123 [22]

Answer:

220 ohms

Explanation:

I = V / R

0.25 = 110 / R

R = 110 / 0.25

R = 440 ohms

Equivalent resistance = 440 ohms

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= 440 / 2

= 220 ohms

8 0
1 year ago
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Why are atoms in a covalent bond usually a certain distance away from each other?
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3 years ago
"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

a = 161.4  kpsi

so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

so here number of cycle N will be  

N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

7 0
3 years ago
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