in December which will have a longer day length a country im notheren hemisphere or southern hemisphere

A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F

ivanzaharov [21]

Answer: $32.65\ m$

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is $\mu_s=0.4$

The coefficient of kinetic friction is $\mu_k=0.2$

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction $\mu_kg=0.2\times 9.8\ m/s^2$

Using the equation of motion $v^2-u^2=2as\\$

insert the values

$0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m$

4 0

3 years ago

A(n)is produced around a wire when an electrical current is in the wire.

Nonamiya [84]

Answer: The correct answer is "magnetic field".

Explanation:

A magnetic field is produced around the current carrying wire.

If you bring compass needle around the current carrying wire then it shows the deflection which indicates that there is a magnetic field around the current carrying wire.

Magnetic fields are the area around the surrounding of magnet in which magnetic force can be experienced.

Therefore, a magnetic field is produced around a wire when an electrical current is in the wire.

5 0

3 years ago

Read 2 more answers

When light is reflected by a mirror, the angle of incidence is always A. equal to the angle of reflection. B. less than the angl

ankoles [38]

When light is reflected by a mirror, the angle of incidence is always <span>A. equal to the angle of reflection. We know this by the Law of Reflection.</span>

6 0

3 years ago

Read 2 more answers

The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from

Allushta [10]

Answer:

y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

let's substitute

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

0 = v₀ sin θ - 9.8 t

Let's write our system of equations

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

v₀ sin θ = 9.8 t

we substitute

10 = (9.8 t) t - ½ 9.8 t2

10 = ½ 9.8 t2

10 = 4.9 t2

t = √ (10 / 4.9)

t = 1,429 s

Now let's use the first equation and the last one

45 = v₀ cos θ t

0 = v₀ sin θ - 9.8 t

9.8 t = v₀ sin θ

45 / t = v₀ cos θ

we divide

9.8t / (45 / t) = tan θ

tan θ = 9.8 t² / 45

θ = tan⁻¹ ( 9.8 t² / 45 )

θ = tan⁻¹ (0.4447)

θ = 24º

Now we can calculate the maximum height

v_y² = $v_{oy}^2$ - 2 g y

vy = 0

y = v_{oy}^2 / 2g

y = (20 sin 24)²/2 9.8

y = 3,376 m

the other angle that gives the same result is

θ‘= 90 - θ

θ' = 90 -24

θ'= 66'

for this angle the maximum height is

y = v_{oy}^2 / 2g

y = (20 sin 66)²/2 9.8

y = 17 m

thisis the correct

6 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

Burka [1]

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

8 0

3 years ago

in December which will have a longer day length a country im notheren hemisphere or southern hemisphere​

in December which will have a longer day length a country im notheren hemisphere or southern hemisphere