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Pavlova-9 [17]
3 years ago
5

A 1100-kg car pulls a boat on a trailer. (Enter the magnitude.) (a) What total force (in N) resists the motion of the car, boat,

and trailer, if the car exerts a 1860-N force on the road and produces an acceleration of 0.600 m/s2
Physics
1 answer:
Finger [1]3 years ago
5 0

Answer:

780 N  

Explanation:

The combined mass of boat and trailer is, m= 700 kg.

Force applied by car, F= 1860 N

acceleration produced on the total system, a = 0.600 m/s²

Apply Newton's second law:

F-F_r=m_na

m_n=700+1100 =1800 kg

Substitute the values:

1860-F_r=1800\times0.6\\F_r=780 N

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Explain resolution of Force​
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Answer:

it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.

Explanation:

7 0
3 years ago
What is the acceleration of a 31 kg object pushed with a force of 42N
Ray Of Light [21]
I belive it could be 6.5 but I could be wrong
3 0
3 years ago
A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that
NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0
3 years ago
A scientist classifies some plants into two groups
gizmo_the_mogwai [7]

Answer:

I'm pretty sure the answer is C . since deciduous trees are trees that loose leaves seasonally and coniferous trees are trees that don't loose leaves seaosnally and survive through the winter.

8 0
3 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
3 years ago
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