PLEASE I NEED HELP!!!

a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the ca

Inga [223]

Answer:

The cart's acceleration is $\approx 3.71\,\,\frac{m}{s^2}$

Explanation:

Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.

Net force = 95.4 N -36.0 N = 59.4 N

Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:

$F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}$

8 0

3 years ago

What is the volume in millilitre of a 250 cm3 beaker

makkiz [27]

Answer:

250 mL

Explanation:

6 0

3 years ago

What are the properties of different types of electromagnetic waves?

Pani-rosa [81]

Answer:

1. All electromagnetic waves travelS at 3×10⁸m/s in vacuum

2.They are transverse in nature.

3.they can travel through vacuum.( doesn't require any materials for their transmission.

4.They under go phenomena of diffraction,reflection, refraction, interference and polarization.

HOPE IT HELPED.

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6 0

2 years ago

At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

andre [41]

Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

4 0

3 years ago

الاضطراب العصبي يحدث بسبب ؟<br><br>

tekilochka [14]

Explanation:

ا: قد يكون وراء هذه الظاهرة ميول وراثية، والتي تتسبب في صدور ردود فعل غير منتظمة في أماكن مختلفة معينة في الدماغ. تكون ردود الفعل هذه مسؤولة عن ظهور العلامات التي تميز العلامات المرافقة التي تظهر عند المرور في ظروف القلق.

ب: ان الفرضية السائدة والأساسية هنا، تقضي بأن العُصاب يظهر كردة فعلِ لفشل المنظومة النفسية الداخلية، في التأقلم مع القلق النابع من معاناة نفسية كبيرة. أو قد تنجم عن فشل المنظومة النفسية الداخلية في التأقلم مع الظروف الرضحية (التراجيدية) الخارجية والمحيطة.

ج:قد تظهر اضطرابات العُصابية من مزج بين تأثير التجارب الرضحية عند الطفولة، مع أسلوب الحياة المعاصر، والذي يتميز بوجود الإجهاد النفسي وبضعف الثقة بالمقومات الشخصية، الاجتماعية والاقتصادية

8 0

4 years ago