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kiruha [24]
3 years ago
13

A box that has a mass of 80 kg slides down a ramp with a 30 degree angle. The free-body diagram shows the forces acting on the b

ox. Ignoring friction and air resistance, what is the acceleration of the box, to the nearest tenth? Answer choices are 0.5 m/s2 4.9 m/s2 8.5 m/s2 9.8 m/s2
Physics
2 answers:
zimovet [89]3 years ago
8 0
The first one I believe
Nonamiya [84]3 years ago
8 0

Answer:

a= 4.9 \frac{m}{s^{2} }

Explanation:

Force equation

F= m*a

F= W*g*sen (30)\\F= 80 kg * 9.8 \frac{m}{s^{2} } * sen (30)\\F= 80 kg * 9.8 \frac{m}{s^{2} } * 0.5\\F= 392  *\frac{kg*m}{s^{2} } = 392 N

a= \frac{F}{m} = \frac{392 N}{80 kg} = \frac{392 \frac{kg*m}{s^{2} } }{80 kg}\\ a= 4.9 \frac{m}{s^{2} }

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An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time
blagie [28]

Answer:

100m

Explanation:

s = ut +  \frac{1}{2} a {t}^{2}

u=0;t=10sec;a=2m/s²

s = 0(10) +  \frac{1}{2}(2 \times  {10}^{2} )

s=10²;100m

7 0
3 years ago
Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil
Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0
3 years ago
· A hot, just-minted copper coin is placed in 101 g of water to cool. The water
Mariana [72]

Answer:

135g

Explanation:

6 0
3 years ago
The state of a medium affect the speed of sound
fenix001 [56]

Answer:

no it can not effect the speed of sound not shure tho

5 0
2 years ago
Your Lead Teaching Assistant is initially seated on the top of a hemispherical ice mound of radius R = 30 m. Approximate the ice
saw5 [17]

Answer:

<em>Height = 5.65 km</em>

Explanation:

3\pi r^2 is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.

So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km

7 0
2 years ago
Read 2 more answers
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