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3 years ago

15

A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc

eleration will be greater for the:

1 answer:

saul85 [17]3 years ago

8 0

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

$F = kx$

now we know by Newton's II law

$F = ma$

so here we have

$a = \frac{F}{m}$

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

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A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot

sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

Initial momentum = Final momentum

4.4v = 12.364 i + 15.092 j

v =2.81 i + 3.43 j

Magnitude

$v=\sqrt{2.81^2+3.43^2}=4.43m/s$

Direction

$\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0$

50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0

3 years ago

What property do liquids and gases share

vodka [1.7K]

Answer:

A is correct

Explanation:

6 0

3 years ago

A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =

Alchen [17]

Answer:

this makes no since so i cant help you here sorry

5 0

1 year ago

Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

4 0

3 years ago

An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,

cupoosta [38]

First overtone of open organ pipe is given as

$f_{1o} = \frac{v}{L_1 + 2e}$

first overtone of closed organ pipe is given as

$f_{1c} = \frac{3v}{4(L_2 + e)}$

now they are in unison so we will have

$\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}$

$\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}$

$90 + 6e = 92 + 4e$

$e = 1 cm$

so end correction of both pipes is e = 1 cm

8 0

3 years ago