Two forces that are not equal in size are

How much heat is needed to raise the temperature of 9g of water by 17oC?

andre [41]

Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7

6 0

2 years ago

A vehicle travels from a 30m marker to a 100m marker. What is the change in distance?

sesenic [268]

The change in distance is 30 because if you subtract both number you'll get 30

3 0

3 years ago

Read 2 more answers

A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

$\theta =44.8{\circ}$

using $v^2-u^2=2as$

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration $(gsin\theta )$

s=distance traveled

$0-(1.72)^2=2(-9.81\times sin(44.8))s$

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

$0=1.72-gsin(44.8)\times t$

$t=\frac{1.72}{9.8\times sin(44.8)}$

t=0.249 s

(c)Speed of the block at bottom

$v^2-u^2=2as$

Here u=0 as it started coming downward

$v^2=2\times gsin(44.8)\times 0.214$

$v=\sqrt{2.985}$

v=1.72 m/s

3 0

2 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

Someone help please.....

Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean) = $\frac{0.125+0.091+0.0625}{3}$

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0

2 years ago