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zaharov [31]
3 years ago
9

What is the mass of an object that has an acceleration of 2.63m/s2 when a unbalanced force of 112N is applied to it?

Physics
1 answer:
exis [7]3 years ago
4 0

Answer:

42.58kg

Explanation:

By Newton's second law, F = ma.

F is the force being applied, in this case 112N. a is the acceleration, in this case 2.63 m/s^2.

Thus, with some simple algebraic manipulation, we get the mass to equal:

m = F/a = 112N / 2.63 m/s^2 = 42.58kg

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A 50.0-g ball traveling at 25.0 m/s bounces off a brick wall and
liubo4ka [24]

Answer:

13,400 m/s²

Explanation:

Average acceleration is the change in velocity over time:

a = Δv / t

a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s

a = 13,400 m/s²

4 0
3 years ago
A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi
bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc    where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

rf = \sqrt{rfX^2+rfY^2} = 245.45km

And the angle measure from the y-axis is:

\alpha =atan(rfX/rfY) = 21.45\°

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0
3 years ago
The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10
Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁  ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0
2 years ago
A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio
xxTIMURxx [149]

The angle of the wedge is 30°.

Answer:

5.88 ft/s

Explanation:

a) The block will slide down due to it's weight.

initial velocity u= 0

final velocity, v

acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

Use third equation of motion:

v^2-u^2 = 2as

substitute the values and solve for v

v^2-0 = 2\times 16 \times 3 =96 ft^2/s^2\\v = 9.8 ft/s

b) Use conservation of momentum:

Initial momentum of the system  = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s

3 0
2 years ago
Ahmad is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s^2 for 5 s to reach a speed of 2.2 m/s. The to
snow_lady [41]

Answer:

1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo

Explanation:

The given parameters are;

The mass of Christian and his bicycle = 54 kg

The mass of the cargo = 12 kg

1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity

∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo.

7 0
2 years ago
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