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2 years ago

15

if force accelerate mass with acceleration and the same force accelerate the mass which is quarter of the first one by

1 answer:

Degger [83]2 years ago

4 0

Answer:

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Explanation:

hdhdbduxnrbdbsnsjjejfndnerf

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A person jogs 4.0 km in 32 minutes, then 2.0 km in 22 minutes, and finally 1.0 km in 16 minutes. What is the jogger's average sp

kupik [55]

Answer:

22

Explanation:

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7 0

3 years ago

A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the

Natali [406]

There are three forces acting on the book.
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.

4 0

3 years ago

Can someone answer questions 1-3 please.

Lunna [17]

Answer:

Question 1 is C.

thats all ik

Explanation:

4 0

3 years ago

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

UkoKoshka [18]

Answer:

a) $W_g =61.25J$

b) $W_k = -46.25J$

c) $W_N = 0$

d) $W_g$ would be the same.

$W_k$ would decrease.

$W_N$ would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

$F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J$

b) The friction force is equal to the normal force times coefficient of friction.

$F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J$

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

$h = x\sin(\theta)$

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of $x\sin(\theta)$ .

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0

3 years ago

A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.

evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

where

(A) a = -k × v

at v= 910 ft/s

a = - 1.987 × 10⁶ ft/s²

(B) at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{dv}{v} = -kdt$

$\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt$

$ln(200.2)-ln(900) = -kt$

t = 6.84 × 10⁻⁴ s

3 0

2 years ago