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3 years ago

The third one pls help

Physics

1 answer:

Nadya [2.5K]3 years ago

7 0

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

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Answer: -200

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2 years ago

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A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac

Charra [1.4K]

Answer:

Time of flight A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight A is greatest .

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3 years ago

Name the type of component that has a greater resistance as the current through it increases

Gnesinka [82]

Answer:

filament bulb, filament lamp

Explanation:

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3 years ago

A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry

Sauron [17]

Answer:

a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

$F_{e}$ - $T_{x}$ = m a

Axis y

$T_{y}$ - W = 0

Initially the system is in equilibrium, so zero acceleration

Fe = $T_{x}$

T_{y} = W

Let us search with trigonometry the components of the tendency

cos θ = T_{y} / T

sin θ = $T_{x}$ / T

T_{y} = cos θ

$T_{x}$ = T sin θ

We replace

q E = T sin θ

mg = T cosθ

a) the electric force is

$F_{e}$ = q E

E = $F_{e}$ / q

E = -0.032 / 80 10⁻⁶

E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

q E / mg = tan θ

θ = tan⁻¹ qE / mg

Let's calculate

θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

θ = tan⁻¹ 0.3265

θ = 18 ⁰

sin 18 = x/0.30

x =0.30 sin 18

x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

$F_{e}$ = m aₓ

aₓ = q E / m

aₓ = 80 10⁻⁶ 4 10² / 0.01

aₓ = 3.2 m / s²

Axis y

W = m $a_{y}$

a_{y} = g

a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

a = √ aₓ² + a_{y}²

a = √ 3.2² + 9.8²

a = 10.31 m / s²

The Angle meet him with trigonometry

tan θ = a_{y} / aₓ

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-9.8) / 3.2

θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0

3 years ago