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3 years ago

8

Please help asap 100 POINTS and BRAINLIEST to best answer.

2 answers:

andrew11 [14]3 years ago

6 0

D. :)

lmk if im right or wrong

borishaifa [10]3 years ago

6 0

Answer:

D. 5ºC

Explanation:

Not B, because 1k= -457.87 fahrenheit. That means it is also not C. Since the icicles are melting, then that means it is not A either.

5ºC=about 41ºF

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a car has a velocity of 10ms-1. it accelerates at 0.2 ms-2 for half minute find the total distance travelled and final velocity

Readme [11.4K]

Answer:

Final velocity = 16 m/s

Total distane travelled = 390 m

Explanation:

We can use equation of motion to solve this:

$v = u + at \\ v = 10 + 0.2(30) \\ v = 16 {ms}^{ - 1}$

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 10(30) + \frac{1}{2} (0.2) {(30)}^{2} \\ s = 390m$

5 0

3 years ago

yulyashka [42]

Answer:

I can msg it toyou

------->------>---->

Explanation:

5 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

4 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Sedbober [7]

Answer:

v = 1.2 m/s

Explanation:

The wavelength of the waves is given as the horizontal distance between the crests:

λ = wavelength = 5.5 m

Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:

T = Time Period = 2(2.3 s) = 4.6 s

Now, the speed of the wave is given as:

$v = f\lambda$

where,

v= speed of wave = ?

f = frequency of wave = $\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz$

Therefore,

$v = (0.217\ Hz)(5.5\ m)\\$

<u>v = 1.2 m/s</u>

5 0

3 years ago

How do I do this?<br> Series and parallel circuits.

Svetlanka [38]

Connect the little light thingys to the top of the battery

8 0

3 years ago