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3 years ago

5

As temperature increases, what happens to the density of ocean water? A. changes unpredictably B. decreases C. does not change D

. increases

1 answer:

Lelu [443]3 years ago

4 0

As the temperature of water increases, the density of water will decrease.

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Tourists covered 255 km for a 4-hour ride by car and a 7-hour ride by train. what is the speed of the train, if it is 5 km/h gre

LekaFEV [45]

The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
$S_c = v_c t_c$
where
$v_c$ is the car's speed
$t_c = 4 h$ is the duration of the car ride

Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
$S_t = v_t t_t$

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
$S=255 km = S_c + S_t$
which becomes
$255 = 4 v_c + 7 v_t$ (1)
we also know that the train speed is 5 km/h greater than the car's speed:
$v_t = 5 + v_c$ (2)

If we put (2) into (1), we find
$255 = 4v_c + 7(5+v_c)$
and if we solve it, we find
$v_c = 20 km/h$
$v_t = 25 km/h$

So, the car speed is 20 km/h and the train speed is 25 km/h.

4 0

3 years ago

Write<br>any<br>to increase<br>magnitude of current in dynamo

Svetach [21]

Answer:

Explanation:

It can be increased by: increasing the rate of rotation. Increasing the strength of the magnetic field. Increasing the number of turns on the coil.

Hope this helps

plz mark it as brainliest!!!!!!

4 0

2 years ago

An over-caffeinated student stands on a table which weighs 600 newtons. The student has a mass of 50 kg. What is the weight of t

liberstina [14]

Answer:50kg

Explanation:

6 0

3 years ago

2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the

Lubov Fominskaja [6]

Answer:

$P=39.2205\, watt$

$E=374.948 \,cal$

Explanation:

Given that:

force applied, $F=66\,N$

displacement, $s=23.77\,m$ (length of a tennis court)

time taken for pushing, t = 40 s

Since, work is given by:

$W=F.s$

$W=66\times 23.77$

$W=1568.82\,J$

Now, power is given as:

$P=\frac{W}{t}$

$P=\frac{1568.82}{40}$

$P=39.2205 \,watt$

Calories consumed is:

$E= 1568.82\times 0.239$

$E=374.948\, cal$

3 0

3 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago