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3 years ago

A body accelerates uniformly from rest at 2m/s square. calculate it's velocity after traveling 9m.

Physics

1 answer:

Talja [164]3 years ago

3 0

Answer:

6m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 2m/s²

Distance = 9m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we use the expression below:

v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

v² = 0² + (2 x 2 x 9) = 36

v = 6m/s

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When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g

mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=m $c_p$ dT

Where m is the mass $c_p$ = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

$T_1$ =25°C

$T_2$ we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25- $T_2$ )

25- $T_2$ =22

$T_2$ =3°C

7 0

3 years ago

Read 2 more answers

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col

DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

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3 years ago

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3 years ago

A force Fx=cx-3.00x2 acts on a virus as the virus moves along an x axis, with F measured in Newtons, x in meters, and c a consta

andrey2020 [161]

Answer:

Explanation:

We are given that

$F_x=cx-3x^2$

K.E at x=0 m=20 J

K.E at x=3 m=11 J

We have to find the value of c.

By work energy theorem

Work done=Change in kinetic energy

W= $\int_{0}^{3} Fdx=\int_{0}^{3}(cx-3x^2)dx$

$W=[\frac{cx^2}{2}-x^3]^{3}_{0}$

$W=\frac{9c}{2}-27$

$\Delta K=11-20=-9 J$

$-9=\frac{9c}{2}-27$

$-9+27=\frac{9c}{2}$

$18=\frac{9c}{2}$

$c=\frac{2}{9}\times 18=4$

4 0

3 years ago

A body accelerates uniformly from rest at 2m/s square. calculate it's velocity after traveling 9m.​

A body accelerates uniformly from rest at 2m/s square. calculate it's velocity after traveling 9m.