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2 years ago

If a car has a mass of 4000 kg and the force applied to it is 10000 N. What is its acceleration?

Physics

1 answer:

Savatey [412]2 years ago

5 0

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{10000}{4000} = 2.5 \\$

We have the final answer as

Hope this helps you

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MATHPHYSSSSSSSS PLEASEEEEEE IM SORRY YOU PROBABLY HATE ME

inysia [295]

Answer:

3.1 m/s

Explanation:

First, find the time it takes for the cat to land. Take down to be positive.

Given:

Δy = 0.61 m

v₀ = 0 m/s

a = 9.81 m/s²

Find: t

Δy = v₀ t + ½ at²

(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²

t = 0.353 s

Now find the horizontal velocity needed to travel 1.1 m in that time.

Given:

Δx = 1.1 m

a = 0 m/s²

t = 0.353 s

Find: v₀

Δx = v₀ t + ½ at²

(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²

v₀ = 3.1 m/s

3 0

2 years ago

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m

Zanzabum

Answer:

$T_{1}$ = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about $T_{1}$ we have:

(M + m) g * 0.5L - $T_{2}$ (L - d) = 0

⇒ $T_{2}$ = [(M + m) g * 0.5L] ÷ (L - d)

$T_{2}$ = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

$T_{2}$ = 59.535 ÷ 2.4

$T_{2}$ = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

$T_{1}$ + $T_{2}$ = W + w ------- Eqn 1

Substituting T2, W & w into the Eqn 1

$T_{1}$ + 24.81 = 26.46 + 13.23

$T_{1}$ = <u>14.88</u> N

8 0

2 years ago

Which of these is NOT an effect of humor?

IgorLugansk [536]

Feelings of jealousy and envy

7 0

3 years ago

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one

alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R= $13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ ohm

Resistance of long piece of wire= $7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire = $\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K= $\frac{\rho}{A}$ =Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'= $\frac{L}{8}$

Length of long piece of wire= $L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of wire= $\frac{\frac{L}{8}}{L}\times 100=12.5%$ %

The resistance of the short piece= $\frac{13}{8}\Omega$

The resistance of the long piece= $\frac{91}{8}\Omega$

8 0

3 years ago

A ball is thrown up in the air. When it falls back down and reaches the ground, it's final downward velocity is 8.0 m/s. What wa

hram777 [196]

Given : Time taken to reach the maximum height t=3 s a=−g=−10m/s
2

The initial velocity of the ball can be calculated by,
Using v=u+at
∴ 0=u−10×3 ⟹u=30 m/s

Using S=ut+
2
1

at
2

∴ S=30×3+
2
1

×(−10)×3
2
=45m

8 0

2 years ago

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