Ask question

Ask question

All categories

3 years ago

10

A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre

ased from 25ºC to 37ºC. How much heat energy did the water absorb?
1: 2,508 joules
2: -2,508 joules
3: 5,225 joules
4: 7,733 joules

1 answer:

butalik [34]3 years ago

6 0

Answer: a) 2,508

Explanation:

You might be interested in

The flow of electrons through a wire or any conductor is called____.

Karo-lina-s [1.5K]

Electricity. Ruler. 69. N.

5 0

2 years ago

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

Natali [406]

centripetal acceleration is given by formula

$a_c = \omega^2*R$

given that

$a_c = 34.1 m/s^2$

$R = 5.91 m$

now we have

$\omega^2 R = 34.1$

$\omega^2 * 5.91 = 34.1$

$\omega^2 = 5.77$

$\omega = 2.4 rad/s$

so the ratationa frequency is given by

$\omega = 2 \pi f$

$2.4 = 2 \pi f$

$f = \frac{2.4}{2\pi}$

$f = 0.38 Hz$

7 0

3 years ago

An electrician timed a box traveling over a conveyor and measured 25 sec to travel 30 feet. If the gearbox ratio is 20:1 and the

nataly862011 [7]

Pppeejsidoocicjddkdkcocfoodoocoie

4 0

3 years ago

What effects do coil of wire on a compass after it has been connected to a battery

Yanka [14]

Explanation:

The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.

8 0

2 years ago

In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo

RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

- The equation for the calculation of the period in a spring-masss system

$T = \sqrt{\frac{m}{k} }$ ( 1 )

- The equation for the velocity of a simple harmonic motion

$x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)$ ( 2 )

where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

$T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s$

and by reeplacing it in ( 2 ):

$v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi sin(9\pi ) = 0$

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.

5 0

3 years ago