All categories

3 years ago

Help me... my science is horrible. I will mark brainliest, would prefer fast answer...

Physics

1 answer:

Elena L [17]3 years ago

7 0

vacuum walls: prevents heat loss by comduction and radiation

silvered walls prevent loss by radiation

ceramic base prevents loss by conduction

cap prevents by radiation and convection and reduces by comduction

You might be interested in

A tennis player standing 12.6 m from the net hits the ball at 3.18° above the horizontal. To clear the net, the ball must rise a

SOVA2 [1]

At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m.
So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY.......
We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION:
v² = u² + 2as
0 = u² - 2gh
u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity)
So rearranging,
velocity = (velocity in y direction only) / sin 3°
= √(2gh)/sin 3°
= (√(2 x 9.8 x 0.33)) / sin 3°
= 49 m/s at 3° to the horizontal

3 0

3 years ago

8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At

Mrac [35]

<u>Answer:</u>

Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

Velocity of car in first hour = 70 mph

Velocity of car in second hour = 60 mph

Velocity of car in third hour = 80 mph

Acceleration = Change in velocity / Time

Acceleration in second hour = (60 - 70)/1 = -10 mph²

Acceleration in third hour = (80 - 60)/1 = 20 mph²

So positive acceleration is in third hour and negative acceleration is in second hour.

8 0

3 years ago

A metal ring 4.30 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

CaHeK987 [17]

Answer:

A)0.00966 N/C

B) counterclockwise direction

Explanation:

We are given;

Diameter of the metal ring; d = 4.3 cm

Radius;r = 2.15 cm = 0.021- m

Initial magnetic field, B = 1.12 T

Rate of decrease of the magnetic field;dB/dt = 0.23 T/s

Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :

∫E•dr = d/dt∫B.A •dA

This gives;

E(2πr) = dB/dt(πr²)

Gives;. 2E = dB/dt(r)

E = dB/dt × 2r

We are given;

E = 0.23 × 2(0.021)

E = 0.00966 N/C

The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C

B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet

6 0

3 years ago

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 965m/s. the mass of the hunter (in

weqwewe [10]

When the gun is fired horizontally :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 m/s

M = mass of the hunter including gun = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (965) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.056 m/s

so recoil velocity comes out to be 0.056 m/s

When the gun is fired at angle 56.0⁰ above the horizontal :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 Cos56 = 539.62 m/s

M = mass of the hunter including gun = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (539.62) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.031 m/s

so recoil velocity comes out to be 0.031 m/s

3 0

3 years ago

An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

Ugo [173]

Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

The initial speed of the object is $u = 8 \ m/s$

The greatest height it reached is $h = 15 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2gH$

At maximum height v = 0 m/s

$0^2 = 8^2 + 2 * - 9.8 * H$

=> $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

$s = h - H$

=> $s = 15 - 3.27$

=> $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

$v = u + (-g) t$

At maximum height v = 0 m/s

$0 = 8 - 9.8t$

=> $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

$h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

$15 = 0* t + 4.9 t^2$

=> $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

$T_t = t + t_1$

=> $T_t = 0.8163 + 1.7496$

=> $T_t = 2.5659 \ s$

6 0

3 years ago