All categories

3 years ago

What can a paleontologist tell from fossil footprints of a dinosaur

Physics

2 answers:

weeeeeb [17]3 years ago

8 0

They can tell what type/species the dinosaur was

AfilCa [17]3 years ago

7 0

Answer:

IDK

Explanation:

Sorry But I not sorry!!!!!!!!!!

LOL:}

You might be interested in

3 small identical balls have charges -3×10^-12 ,8×10^-12 and 4×10^-12 find the charge

Norma-Jean [14]

Answer:

3×10^-12 C

Explanation:

The total of the three charges is ...

(-3 +8 +4)×10^-12 C = 9×10^-12 C

Assuming the charge is equally distributed between the balls when they are brought in contact, the charge on each ball will be ...

(9/3)×10^-12 C = 3×10^-12 C

7 0

3 years ago

during a workout, a football player pushes a blocking dummy a distance of 30 m. while pushing the dummy the same distance a seco

babymother [125]

$Power=\frac{Work}{Time}=\frac{Force\times distance}{time}$

If he wants to increase power, force must increase and decrease time.

8 0

4 years ago

Read 2 more answers

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago

An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici

Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539 J

From first law of thermodynamics

Q₁ = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

$\eta=\dfrac{W}{Q_1}$

$\eta=\dfrac{1210}{1749}$

$\eta=0.6918$

η = 69.18 %

We know that rate of work done is known as power

$P=\dfrac{W}{t}$

$P=\dfrac{1210}{0.305}\ W$

P=3967.21 W

8 0

3 years ago

What is the tendency of matter to resist any change in motion?

salantis [7]

Answer: inertia

Explanation:

Inertia is the tendency of an object to resist changes in its state of motion

6 0

3 years ago