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kap26 [50]
2 years ago
7

A 100kg car starts from rest at the top of a hill with a height of 50m. which of the following is this cars kinetic energy at th

e bottom of the hill?
98,000 j
5000 j
49,000 j
150 j​
Physics
1 answer:
Anarel [89]2 years ago
8 0

Explanation:

KE =1/2 MV^2

HOPE YOU COULD SOLVE FROM NOW..

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3 years ago
A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
Urgent if earths gravitational field strength is 10 N/kg how many newtons does a 30kg object weigh
egoroff_w [7]

weight of 30 Kg object= 300 N

Explanation:

Weight= m I

m= mass=30 kg

I= gravitational filed strength= 10 N/kg

weight=30 (10)

weight=300 N

5 0
2 years ago
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\&#10;g_0=g'\\&#10;g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\&#10;ag_0=g_0-w^2r\\&#10;w^2r=g_0(a-1)&#10;
Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
3 years ago
How many butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitation
Mazyrski [523]

Answer:

175.96 g

Explanation:

Potential energy required for the man to climb 7.07 km = m g h.

= 64 x 9.8 x 7070

= 4.434 x 10⁶ J

= 4.434 X 10⁶ / 4.2 cals

= 1.0557 x 10⁶ cals

= 1.0557 x 10⁶ / 6000 g of butter

= 175.96 g of butter.

3 0
2 years ago
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