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navik [9.2K]
3 years ago
8

Find acceleration. Will give brainliest!

Physics
2 answers:
Art [367]3 years ago
8 0

Answer:

\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}

Explanation:

\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}

\displaystyle \mathrm{a = \frac{v - u}{t}}

\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}

\displaystyle \mathrm{a = \frac{45- 0}{10}}

\displaystyle \mathrm{a = \frac{45}{10}}

\displaystyle \mathrm{a = 4.5}

\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }

zalisa [80]3 years ago
7 0

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

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Mariulka [41]

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration a_c, so:

a_c = w^2r

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

a_c = 2.6m/s2

So, Replacing the data, and solving for w, we get:

2.6m/s = w^2(60.5m)

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2\pi)= 1.976 rpm

3 0
3 years ago
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muminat

Answer:

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2 years ago
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A ball on a cart is moving at a rate of 2 m/s. The cart suddenly stops and the ball continues to travel in the same direction at
sukhopar [10]

Answer:

The first

Explanation:

Cause it will continue in motion till another force is applied

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3 years ago
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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
A brand new corvette can go from 0 to 85 miles per hour in 4.8 seconds.
klasskru [66]

Answer:

(A) 7.9 m/s^{2}  

(B) 19 m/s

(C) 91 m

Explanation:

initial velocity (U) = 0 mph = 0 m/s

final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s

initial time (ti) = 0 s

final time (t) = 4.8 s

(A) acceleration = \frac{V-U}{t}

         = \frac{38-0}{4.8} = 7.9 m/s^{2}  

(B) average velocity = \frac{V+U}{2}

     =\frac{38+0}{2} = 19 m/s

(C) distance travelled (S) = ut + 0.5at^{2}

  =  (0 x 4.8) + 0.5 x 7.9 x 4.8^{2} = 91 m

5 0
3 years ago
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