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3 years ago

14

In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35degree angle. Compressing the spring 18 cm ca

uses the ball to hit the floor 1.5m below the point at which it leaves the spring after traveling 6.0m horizontally.

1 answer:

olasank [31]3 years ago

4 0

spring constant K = 45.5412N/m

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Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of

kolbaska11 [484]

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

7 0

3 years ago

What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is doubl

vlabodo [156]

Answer:the force will remain same

Explanation:

because force is equal to the ratio of magnitude and distance

4 0

3 years ago

Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC

scoray [572]

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

$m = 0.0250 kg$

so the energy released in form of vapor is given as

$Q = mL$

$Q = (0.0250)(2.25 \times 10^6)$

$Q = 56511 J$

now the heat required by remaining water to bring it from 15 degree to 100 degree

$Q_2 = ms\Delta T$

$Q_2 = (4 - 0.025)(4186)(100 - 15)$

$Q_2 = 1.41\times 10^6J$

total heat required for above conversion

$Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J$

now by heat energy balance

heat given by granite = heat absorbed by water

$m(790)(500 - 100) = 1.47 \times 10^6$

$m = 4.65 kg$

4 0

4 years ago

What is unit of pressure

REY [17]

Answer:

pascal

Explanation:

5 0

3 years ago

Read 2 more answers

hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward

dimulka [17.4K]

For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

t = 5 m / 6.65 m/s

t = 0.75 s

Result:

The time will be <u>0.75 seconds.</u>

7 0

3 years ago

Read 2 more answers