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ANEK [815]
3 years ago
7

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di

agram shows the forces acting on the cart. The mass of the cart, to the nearest whole number, is _______  kg
Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

The cart accelerates at 1.4 m/s² horizontally.

Horizontal force = Fcosθ = 250 cos35° = 204.79N

We have F = ma

Substituting

        204.79 = m x 1.4  

              m = 146.28 kg = 146 kg

Mass of the cart = 146 kg

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Alecsey [184]

Yes, light enters your eye and makes things visible. It's only a small portion of the elctromagnetic spectrum. It's called "visible light." What other things does the question want to know? I could go in a describe what light really is but I think itd be unnecessary.

6 0
3 years ago
Calculate the potentia energy of a car with a mass of 3800kg that is on a hill 110 meters above sea level? USE 10 instead of 9.8
Arada [10]

Potential energy (PE ) = m g h

Where:

m = mass = 3800 kg

g = acceleration due gravity = 10 m/s^2

h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

7 0
1 year ago
The mass of the car?
Nonamiya [84]

Answer:

1050 kg

Explanation:

The formula for kinetic energy is:

KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.

We can now substitute the values we know into this equation.

KE = 472 500 J and v = 30 m/s:

472 500 = 1/2 × m × 30²

Next, we can rearrange the equation to make m the subject and solve for m:

m = 472 500 ÷ (1/2 × 30²)

m = 472 500 ÷ 450

m = 1050 kg

Hope this helps!

3 0
2 years ago
Read 2 more answers
Please help me guys never mind the calculations ​
vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

5.2) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

I hope I helped you^_^

7 0
3 years ago
A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi
Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0
3 years ago
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